Buon giorno! We're here in Pisa, Italy, where Galileo Galilei is rumored to have completed some of this famous experiments on motion including dropping objects from the leaning tower. In this unit, we're going to complete some of those experiments ourselves, and we're going to derive the laws of motion that Galileo Galilei figured out in the 16th century.
Before we start answering the question of "What is motion?" or "How do we describe it?", let's quickly go over why we want to learn about motion at all. Well, first, understanding motion gives you an incredible predictive power. Whether you want to know the path that a soccer ball takes as it flies to the goal or the speed of the rollercoaster as it goes through a loop, you will need to know about motion. Second, this is an important tool to have in your physic tool belt. The math and the physical techniques we used will be used throughout the class. Finally, and this is the most important reason for me personally, is to solve one of the nature's puzzles. There're so many facets of nature out there that are just begging to be understood, and motion is one of them. Galileo Galilei did a lot to help us learn about motion, but before we jump into his contributions, let's talked a little bit about Aristotle, whose concept of motion was widely accepted until Galileo came on the scene. To get some of these historical contexts, let's go to Florence, Italy, where we'll talk to Professor Thomas Settle, an expert in the field of Galileo, about some of this context.
He, like a lot of other people in that period—1560s or 1570s— had already come to question a number of the Aristotlean assumptions about things. Now according to the doctrine bodies fell with a speed according to their weight— this is where Galileo starts. Then he began wondering about, well, how do actually bodies move and can we tell. I used to, in front of my class, take a couple of balls and drop them in front of the class. And I'd ask the members of the class, "What do you see?" And most of the students coming from engineering or science said, "We see ball or bodies accelerating in free fall." I would come back at them eventually and say, "Now, I didn't ask you what you were told you were ought to see. What do you actually see?" And if you try it, you'll find that you don't see much. You see the hands holding the ball. You see the hands opening. And you hear a thump or a couple of thumps on the floor, but you don't see anything. And this in part is one of the points that Galileo starts.
According to Aristotle, objects all fall at a constant speed, and that speed depends on their weight, and when Aristotle says "constant speed," what he means is that the instant you release the ball, it's already traveling at the same speed that it will hit the ground. Okay. So, objects fall at constant speed and that speed is proportional to the mass. Now, what does this mean—speed proportional to mass? It means that if you double the mass of an object you double the speed. If you triple the mass, you triple the speed. We can write this proportionality with this funny-looking symbol here. In fact, she looks a little bit like the alpha we saw in the first unit. Now, we can use this proportional relationships to make mathematical calculations. Let's say we have two objects, a one-kilogram and a two-kilogram mass. If this falls with a certain speed S1, this one will follow the speed S2. The way we can use this proportionality relationship to a make calculation is in the following line. We say that the ratio of speed S1 to S2 is equal to the ratio of the masses, and we can say that because they are proportional. In this simple example, I just want you to tell me, if S1 is 2 meters per second and these are the masses, what is S2? Well, we can just plug in here.
S1 and m1 are 2 m/s and 1 kg, s2 is unknown, and m2 is 2 kg. Well, we can just solve for s2. When we carry out the Algebra, we find it's equal to 4 m/s, just like we'd expect--double the mass, double the speed.
With these two completely different beliefs--Aristotle who says that objects fall at a constant speed, and Galileo saying that objects gain speed as they fall, how do we reconcile these different opinions? Well, we're scientists. We look at the evidence. Let's look at two pieces of evidence in particular. First, we have this idea that a mass dropped from a greater height leaves a greater impact, and you could try this yourself. Go outside and drop a brick from just barely off the ground and then drop it again from higher off the ground. You'll see that it'll make a bigger dent if you drop it from higher. Now, Professor Settle mentioned earlier how difficult it is to actually see anything when you drop an object. Well, that's true except for when you drop something like a feather, and when you do that, you can see that it really does tend to just drift down at more or less a constant speed. Now, for each of these points, I want you to select one of these buttons. Do this piece of evidence work in favor of Aristotle's belief or Galileo's and likewise for this one so you should check two of these buttons.
Well, this first point seems to point to Galileo, but the second—feathers fall at constant speed— well, that's exactly what Aristotle said—we are in a conundrum. Galileo has already looked at the evidence and he's still confused. That leaves him only one course of action—he has to do experiments.
Now, Galileo had his work cut out for him. The idea that objects fall at a constant speed had been around for almost 2,000 years by the time Galileo came around. This idea had a lot of, as we say in physics, inertia. It is going to be very difficult to change. If we were able to get the fact that feathers fall at constant speed, Galileo had a pretty difficult task ahead of him. Now Galileo had a proposal for why feathers do actually fall at constant speed. He felt that some resistance provided by the air would actually slow objects down. And for lightweight objects like feathers, this has a huge effect and causes them to fall at more or less a constant speed. Now, the rumors that Galileo's experiments involved him dropping objects off of the leaning tower of Pisa— it's quite possible that he did but it seems unlikely that he would have gotten good results since, as we talked about before, the fall motion is just too fast. Which brings us to the question--how can we slow motion down? This seems like an essential aspect to doing any meaningful experiment with motion. So, should we drop objects through water or some other liquid to slow them down? Should we roll objects at inclined planes? An inclined plane is just another word for a ramp. Or should we still drop our objects through the air but just attached a parachute to them to slow them down? I want you to tell me the best answer and there is only one here—so this may require some thought. Keep in mind some of the points made over here when you're answering this question.
The best answer is to roll objects down inclined planes. The reason why this isn't a great answer is because attaching a parachute is essentially doing the same thing as we do when we drop a feather. We're just making a big bag to catch that air and to slow the object down artificially. We're not observing its somehow natural motion. Same thing when we drop an object through water or another liquid. That water is basically doing the same thing as the air would be doing when we drop a feather, but it's having an even greater effect. If we want to make air resistance negligible, the thing to do is to roll objects down inclined planes. Professor Settle, what do you have to say about this? He eventually decides to try rolling balls down an inclined plane. His first experiment, so-called, on an inclined plane, he is still looking for a proposed uniform motion. What he begins to find is that he is never getting uniform motion. What he's getting always is acceleration. So he's finding acceleration. It seems that this experiment is actually making some progress in disproving Aristotle. You may be saying--big deal, of course objects accelerate when they fall, but this was not common knowledge at the time.
Now, you may be wondering what can we learn from rolling balls down the hill. I mean we all know what it looks like when a ball rolls down the hill right? Well, let's see. This quiz is designed to just test your intuition, and I confessed I would not have passed this test before I actually learned how to mathematically quantify motion. So, let's say we have a ball, and we released it, and at the same instant that we released it we start the timer. So, the timer begins at zero seconds and we release it at this position here. So, after 1 second, it's rolled this far. So, given that we now have ball that's released from rest here and after 1 second is here so I want to know where will the ball be at t = 2 seconds. It's gone from here to here in 1 second. After 2 seconds, do you think it will be here or here or maybe it will be all the way over here. Make your guess by selecting the appropriate button.
After 3 seconds, the ball will be right here. Well, if you got those result, I am very impressed. You have an amazing intuition about the motion of objects. If, like me, you didn't get this question right the first time you tried, let's look at this as a motivation for why we want to quantify motion. Before we jump into the nitty-gritty of quantifying motion, we should probably agree on some terms first.
Let’s begin to answer this question, “What is motion?” Well, let’s say motion is changing position. If something’s position or location is staying the same, well, it’s not moving. So, position is an important concept, important enough to be written in red. Actually, it’s not just position that’s important, it is this idea of changing position which actually leads us to the next concept, velocity or speed, which measures how quickly an object’s position changes over time. So, for example, let’s say we have two objects, A and B, and this one’s position changes from here to here and object B’s position changes from here to here. I want to know which object move faster. Was it object A, object B, or can we not tell from the information I’ve given you? Think about this one, it may not be as obvious as it seems at first.
Well, object B didt move further, but from the information we gave you, we actually can't tell which one moved faster. If it took object B 10 years to move over here and object A only took 10 seconds to move over here, well, clearly object A must have been moving quite a bit faster than object B. So this motion of the years, seconds seems to be playing into motion a lot. We should change our definition. Let's call it changing position over time.
We’re going to be talking about these quantities so much, let’s give them some abbreviations. Position, I’m going to abbreviate with the variable X. And velocity, I'll use the variable V. Now, we just said that measuring an object’s velocity is somehow measuring how fast its position is changing. Well, we can write that with an equation. This equation says that velocity is equal to the change in X over the change in time. The change in position over the change in time. This triangle here is the Greek letter delta, and it represents change. X represents position. This mathematical definition really ties in with our intuitive definition of what velocity means. Before we go any further, let's also talk about how we're going to measure these quantities. What units are we going to use? To measure position, we're going to use the unit meter, which we'll abbreviate to M. While velocity is just position, which is meters divided by some unit of time. And the unit of time we'd like to use is the second. So second. Meters per second. If we were Aristotle, we would stop here because if an object falls with constant velocity - And velocity never changes. We don't really care about anything else. Position and velocity seemed to be enough to quantify motion. But we're talking about Galileo and for him speed is constantly changing. This concept of changing speed is so important we give it its own word, acceleration. And we abbreviate that with A. And acceleration is a measure of how an object's velocity changes with time. Just as velocity measure an object's positional change with time. Using this idea that acceleration is change in velocity, can you complete this equation which defines acceleration?
And the answer is--well, acceleration is change in velocity divided by some elapsed time or change in time.
These equations are definitions of velocity and acceleration, but we can also use them to make some calculations. For example, let’s suppose I drop a ball 5 meters and it takes 1 second for this ball to fall. For example, let's say I drop a ball 5 meters and I measured that it takes 1 second for the ball to hit the ground. What was the ball’s average velocity? And we will talk more about more why I’m using the word “average” later. Δx is change in position, how much is ball’s position changing? Well, it's changing by 5 meters. V equals change in position--5 meters--over change in time--well, it took 1 second. So, 5/1 is just 5 meters per second. Not bad. Now, how about you do one. Let’s say that Galileo drops a ball at the Leaning Tower of Pisa and he finds that after 2 seconds, it’s speed is 20 meters per second. I want you to tell me what was the ball’s acceleration.
Now, there's a few answers you might have given to answer this question. Let's talk about them. First, you might have used this equation, this definition of acceleration, to say okay well the ball changes its velocity by 20 meters per second by going from 0 to 20, which means Δv is equal to 20 m/sec and since it takes 2 seconds to drop Δt is equal to 2 seconds. Well, this is just the equation for acceleration and so the answer is That's your first answer you may have given, maybe 10 m/sec squared. That's probably the most correct, but your second answer, which I would have been very impressed with as well is how did Galileo measure speed. This was hundred of years ago. They had no good way of measuring speed. And so the very question that I gave to you was pretty flawed in that sense. The third answer you may have given is how did he measure time. Galileo did not have any stop watches or fancy chronometers in his day. Let's see what Professor Settle has to say. To realize good results from the inclined plane needed to be able to time precise interval segments of motion down the inclined plane. He did not have chronometers--he used a pot of water with a hole in the bottom which he could fill to a certain level and which he could allow water to flow through which he could open and stop, collecting the water in a container, and then weighing the water.
Let's go back to the experiment. We know he used an inclined plane to roll balls for the purpose of slowing down the motion. We know he used a water clock to measure time. This water clock couldn't measure seconds. He didn't know that the time this clock measured was proportion of the amount of water collected in this collecting device, so he collected this much water in your first experiment and twice as much as water in your second experiment. The second experiment took twice this long. So how they did do the actual experiment? First, you'll roll the ball for 1 unit of time, and a unit here is to find some specified amount of water. You'll let the ball roll and then measure the distance. So maybe after 1 unit of time, the ball had gone from here to here. So how did Galileo do his experiment? Well, you'll have the ball to roll for 1 unit of time or unit of time with a specified amount of water, and you'll measure the distance. This distance you would call 1 unit of distance let say. So if the ball rolled from here to here in 1 unit of time, this became 1 unit of distance. Notice that Galileo wasn't really concerned whether these units were seconds or minutes or meters or inches. He was looking for proportions and improportions. These units always cancel out. So, Galileo really didn't care about these things. After he did that for 1 unit of time, he would allow the ball to roll for 2 units of time and then measure the distance and again for 3 and 4 and so on. He'll keep doing this for 3 units of time, 4, 5, and so on, and when he was done, he would repeat the entire experiment many many times, because more data means you can reduce your error, and we'll talk about that now.
Here's an example of what Galileo's data may have looked like. In this version of the experiment, he ran through the experiment five times. On the first run through when he led the ball to roll 1 unit of time, he measured the distance of 1 unit. After 2 units of time, he got 4.03; 3 was 9.07; and 4 was 16.18. Now he was a good scientist so of course he repeated this process multiple times, and this is data he got in each of those following repetitions. Now with all these wonderful data now he has to analyze it. But how does he take advantage of the fact that he has collected each piece of data multiple times? Presumably, he wants to find patterns in this data but how does he use all of it? He has collected it. He should use it all. Well, the answer is he takes the average. And so a quick review on averages in case you've forgotten or if you don't know. To calculate an average--for example if I want to calculate the average distance ruled in each of these units of time, and I've indicated the average of distance by a D with a bar over it, and in physics a bar over a letter generally means the average. What I'm going to do is add up all my data and divide by the number of data points. Here I would add up 1.00 + 1.01 + 0.99 + 0.99 + 1.01. Divide it by 5 because that's how many data points I have--1, 2, 3, 4, 5. And when I calculate that, well I get an average of 1.00. Can you calculate the average for the other 3 time measurements? Enter your answers in these three boxes.
To calculate these answers, I added up all of the entries in each row. For example, in this row, I added up these numbers divided by 5, and they give me a result of 4.00 here, 9.00 here, and 16.02. I’ve rounded here for this last one. Now, look at this data, isn’t this beautiful? 1, 4, 9, 16.02 though. Well, that 0.02 is sort of a bummer but let’s put it was just 16 per seconds and see if there is any pattern. Well 1, 4, 9, 16, I do see a pattern there. These are squares. One unit, 1 unit of time squared is 1, 2 squared is 4, 3 squared is 9, and 4 squared is 16--that’s an incredible pattern. It seems like there may be some relationship between distance and square of time, but the 16.02 is really bumming me out. Does it really mean that this pattern is false? Well, not necessarily because one thing we forgotten to do so far is calculate error. In the previous scene, we talked about how to calculate the error of a single measurement. May be that measurement measuring the length of the shadow or something. But now, we have multiple measurements with five measurements for each of these calculated values. How do we calculate the error associated with multiple measurements?
When we calculate error with multiple measurements, we're focused on precision. Let's talk about what precision is. Let's say this is you. You're a pretty tall person, you know you are about 2 m tall but you're really curious about exactly how tall you are, so you ask your doctor, Doctor, how tall am I?" She says, "Well, your height is 2.00 meters. You can trust me; I'm a doctor." "Okay," you say, "Can you do that measurement again just so I can feel confident about the measurement?" "Sure, no problem." She does the measurement again. She finds, "Well, you have a height of 1.20 meters. "I did the measurement and I'm a doctor. You can trust me. " You're a little confused so you say, "Well, why don't you make that measurement again, I just want to be sure." "Well, I did the measurement a third time like you asked, and I've found your height. It is 2.80 meters.You could trust me. I am a doctor." Now my question--do you trust your doctor?
I sincerely hope you said no, but why? If you calculate the average of these three numbers, if you add them all up and divide by 3, you find an average height of 2.00 meters. That's about what you expected. This is a pretty good number, isn't it?
Let's pretend you go to see another doctor after this and this doctor is a little more competent. She has purple hair, and when you asked the purple-haired doctor to repeat the measurement three times, she gives you the following numbers for your height. If you calculate the average, you find an average height of 1.97 meters. Now my question is--what's a better estimate of your height? Is it the 2.00 meters measured by the green doctor or the 1.97 meters measured by the purple-haired doctor?
The answer, even though you came in thinking you're about 2 meters tall, is definitely 1.97 meters. Since these numbers are all so much closer to the average value of 1.97, I have more trust in this doctor's ability to make measurements compared to this where the numbers are really far from 2. How do I quantify that trust in a measurement? We've already talked about taking an average. Let's take what's called an average deviation, and this number measures on average how far each is your data points is away from the mean. So let's do an example.
For the first doctor, we have three data points. Data point one, well, it's observed value was 2.00 and the mean was 2.00. It’s actually a distance of 0 from the mean. The absolute deviation of this number, this observation, is 0. What about this guy, 1.2 meters? Well, 1.2 - 2 is equal to 0.8, -0.8. But since I don't really care whether they're overestimating or underestimating, I’m going to always make these numbers positive which I do by taking the absolute value. So, just call that + 0.8. This third measurement of 2.8, I get again a deviation of 0.8. Now, for my three measurements, I have deviation of 0, 0.8, and 0.8. That’s the deviation part. What about the average? Well, you know how to do an average, you take the sum, 0 + 0.8 + 0.8 gives me 1.6 and we divide by a number of data points, so 1.6/3 is equal to about 0.53. So, the average deviation for this first doctor was 0.53 and that unit there is meters. We interpret this by saying when this doctor makes a measurement, I can only say with confidence that she's probably within 0.53 meters but even then that's not always true. In fact, into these measurements, she was outside of that range. The way we would write this is by saying the green-haired doctor measured my height to be 2.00 meters plus or minus 0.53 meters. Now, normal height within half a meter is not so helpful, so let’s see what the purple-haired doctor said. Can you tell me how the purple-haired doctor would state my height including the average and the average deviation? Enter your answer here.
Well, the average is just 1.97, the same average as before, and the deviation-- well, we add up how far each of these deviate from 1.97. This one deviates by 0, 0.01 and also a 0.01 because remember we take absolute values. Now, if we take the average of these three numbers, we get about 0.01 and again, I've rounded up from 0.0066667, but this is clearly a much smaller error and I trust this number The lesson here--beware of green-haired doctors.
With the Galileo's data, we know the average isn't the only thing that matters. We also have to look at the average deviation. I want you to go ahead and calculate the average deviation for each of these four rows. You can go ahead and round to two decimal places.
Now, when we look back at Galileo's data, we had this pattern that we wanted to see emerging--this 1, 4, 9, 16. These seem to be squares, perfect squares. Now, one thing I should mention is an assumption we're making. The assumption we're making when we do this experiment or just about any experiment is that nature is repeatable. By that I mean when we roll the ball for 1 unit of time, if we actually role for 1 unit, the ball will go the exact same distance in all five of these trials. The reason why these numbers aren't all identical is because we either miss-timed or miss-measured the distance. It's an error on our end. It's an error on our end, not on nature's end. To quantify the error on our end, how good we are at making these observations, we've calculated this average deviation. What this average deviation tells us is that for each of these trials, each of these four different experiments, we believe the true distance to be, well, for example, 1 ± 0.01 distance units-- whatever those distances were. That means the true value here could have been between 0.99 and 1.01. That means we believe that the minimum distance this ball actually rolled was 0.99 and the max was 1.01. This is 1 - 0.01, and this is 1 + 0.01. What would be the minimum and maximum possible true distances for this example?
Well, 16.02 - 0.09 is 15.93, and if we add 0.09, we get 16.11. Notice that the number 16.00 still falls within this realm, and so our square relationship that we were looking for before is still entirely possible. It's still, as we would say, within our experimental error.
We have this wonderful relationship. When a ball rolls down a hill, the distance it rolls, and I tried to mark off equal distances here--excuse my terrible drawing-- but the distance the ball rolls is somehow proportional to the square of the time. I say proportional to and not equals because remember we don't know the units of time or distance, in fact. Time was measured by allowing water to flow and then weighing the resulting water. One distance unit was just defined as the distance a ball rolled in one time unit. Okay, so we have this relationship. What can we do with it? Well, it reminds me a little bit of what we saw from Aristotle earlier in the unit. Aristotle thought that all objects fell at constant velocity and that that velocity was proportional to the mass. The way we use this relationship was to set up ratios. If we have two objects, 1 and 2, with masses 1 and 2, and some velocities that we're assuming are proportional to that mass, we can set up this equation. If we happen to know three of these quantities--maybe two masses and one velocity-- we can figure out the unknown quantity. Okay, not so bad. But what is this square do? How does that change the calculation? Well, initially this setup looks just the same as before. The only difference is now we have to square this side--pretty straightforward. Let's practice. We have two identical balls, which we'll number 1 and 2. I let this first ball fall for 2 seconds, and it falls a distance of 20 m. If I let this ball fall for 7 seconds, how many meters will it fall? You can use this equation and enter your answer here.
Well, let's set up our ratio—d₁/d₂. Well, that'd be 20 over—well, let's call this x for now. And then that's equal to t₁/t₂ or (2/7)², and (2/7)² is just 2²/7² or 4/49. Solving this for x, I get 980/4, which is 245 m. We were able to solve this problem using only ratios and our knowledge of the relationship between distance and time for a free falling object.
Now, solving problems using ratios and geometry is something that was very popular in the time of Galileo. Today, it's more common to see equations. Isn't that right, Professor? For Galileo, mathematics was geometry. You find, for instance, the law of free fall expressed not in an equation that we're familiar with but in terms of ratios. This is to this as that is to that. So, let's review what we've learned about motion. We've learned what position, velocity, and acceleration are, and these are the key words we use to describe motion. We've defined them in terms of each other. For example, velocity is defined in terms of a change in position. We've given units to position and velocity. We'll get there for acceleration. And now we've found something truly incredible. A mathematical relationship between time and position for an object in free fall. Now, put yourself in Galileo's shoes. Imagine being the first person to realize that nature not only behaves predictably, but in a way that we can describe using mathematics. It's truly a beautiful thing. From this Galileo actually inferred that acceleration is constant. What Galileo meant by "constant acceleration" was that no matter where you were in the world, whether you were in Pisa or Peru, if you drop an object, it will fall at a constant rate. The acceleration will be the same everywhere and for all objects. Now, to better understand this idea, let's talk about the units of acceleration a little bit more. Well, the units are meters per second per second. Notice that these units include both a time interval—a second— and a speed—meters per second. This sort of hints at the definition for constant acceleration, which just means an object gains equal speed during equal time intervals. If we drop an object, and it starts at rest, and after 1 second it has a speed of 10 m/s, meaning that it's gained 10 m/s of speed in that first second, what will its speed be after 2 seconds? This can be a subtle point, so don't worry if you don't get it right on your first try.
The answer is 20 m/s. It went from 0 to 10 in its first second. It'll go from 10 to 20 in it's second second. That is what constant acceleration means. In fact, after 9 seconds, for example, the object will be going 90 m/s. It seems like, again, we're seeing this proportionality. In fact, this velocity seems to be proportional to time. Here's it's just 10 times whatever the time is. So, we have this relationship--v proportional to t. Wow, we've really discovered a lot about motion.
To make sure we understand this idea, velocity is proportional to time, let's look at an example. Let's imagine that we drop a ball. The instant we drop the ball t = 0. The ball's speed is 0 m/s. If I wanted to represent this for the picture, I would just draw a stationary ball. If my acceleration is 10 (m/s)/s, which we can also say 10 m/s², then after 1 second I will have gained 10 m/s, and I can represent this velocity with a downward arrow. The length of this arrow somehow represents the size or magnitude of this velocity. After another second, this speed has increased to 20, and now my arrow is twice as long. I want you to tell me two things about what happens after 3 seconds. First, the speed and second, where should I put the end of this arrow, which represents that speed? Should I put it here, here, here, or here? Well, the speed is 30. We've just gained another 10 m/s. The arrow should end here. It's length has increased by one more unit. Now, these arrows are something we're going to use a lot in physics. We're going to call them vectors. Notice that these vectors have two properties, a length and a direction. These vectors happen to be pointing downwards, but we'll see soon vectors that point in other directions.
We've really learned a lot. Surely, we'll be able to answer a simple question, right? For example, if Galileo drops a ball off the leaning tower, how far does it fall in 2 seconds? Can we answer that? Well, we want to know distance. I'm thinking we going to use our relationship with x, and it's somehow proportional to t². But unfortunately, this is not enough information. Remember how we solved this. We used ratios. Ratios mean we need to compare two things. If I knew how far the ball fell in 1 second or in 10 seconds or in 7 seconds, yes, I could solve this. But I haven't even told you that. It looks like we might be in trouble. What we want is not a proportionality relationship. We want an equality. We want x equals some constant--some number that never changes--times t². This constant is called the constant of proportionality. Before I just tell you what the constant is let's learn about an amazing physical tool called dimensional analysis where "dimensional" refers to the dimensions or units of a number, because right now we know nothing about this constant. It could have units of length. It could have units of time. It could have units of time squared over length cubed. We just don't know. Let's use dimensional analysis to figure out what the units of this constant should be. We do dimensional analysis by converting everything into units of length, time, and mass, combination of these. Let me show you what I mean. Let's look at this equation x equals something times time squared. Just keep track of the units. Well, x is a unit of length. It's a unit of distance. On the left hand side I'll write length. That equals some unknown here multiplied by something with units of time squared. Solving for my question mark, I find that the units of this unknown constant are length/time squared. What has units of length/time squared? Well, length/time squared could be something like meters per second squared. That's acceleration. This constant of proportionality that relates time and position has units of acceleration. Pretty cool. It turns out that this constant of proportionality is just 1/2 the acceleration of the object that we're talking about. For objects in free fall on earth, this acceleration happens to be equal to 10 m/s², which is why I've been using this as an example acceleration so much. In fact, this number is so important we give it the abbreviation "g." But keep in mind we only use this number when an object's in free fall. When an object's rolling, this acceleration will be--well, you tell me. Does an object rolling accelerate faster or slower?
It accelerates slower. That's the reason we rolled balls down inclined planes to begin with. Galileo needed to slow down motion. Now that we have an equation, we can really solve some interesting motion problems, but we can't solve them all just yet. For example, this equation only describes an object that starts at rest. What if Galileo decided to throw this ball off the leaning tower? Or what if he decided to throw it at an angle. Well, now we're in the realm of 2-dimensional motion. In the rest of this unit we'll talk about how to solve these problems.
Okay, so you have this amazing new skill, and you can calculate all sorts of motion problems. Now, can you calculate every motion problem? The answer is not quite yet. There are two types of problems we still can't solve. We can't solve problems where an object starts with some initial velocity, and we can't solve problems where the object moves in two dimensions, but we'll get to that later. For now let's talk about this initial velocity idea. The variable we're going to use to talk about initial velocity is V with the subscript 0. This comes from the idea that when we start a problem we usually start our timer at 0. So this is at time equals 0 what is the velocity? Let's think back to Galileo standing on top of the leaning tower of Pisa. This time around, instead of just having him drop the ball, let's imagine he could throw the ball straight down with different initial velocities. Now, just to get you thinking, I want to think about this question, when will the ball go farther? If he drops the ball with an initial speed of 0--so that's truly just dropping the ball and not throwing it at all--will the ball go farther then? Let's say in a single second? Or will the ball go farther if he throws it at 10 m/s straight down? In one second, when will the ball go farther?
The answer is if you give the ball some initial velocity, it will go farther in that first second of travel. Okay. But how much farther is always the question in physics. How much? Let's quantify.
So, you already know how to answer this question when the V₀ is 0. That is, he's dropping the ball, not throwing it, actually. In fact, I think you can answer this using our equation x = ½gt² where we remember that g is about equal to 10 m/s². Once you go ahead and calculate the distance for these three cases. If I plug in 10 for g and 1 for t, I get ½ * 10 * 1, which is 5 m. Plugging in 2, I get 20 m, and plugging in 3, I get 45 m. Now, if Galileo could have taken an accurate velocity, time, and distance measurements, and he did the following experiment with different values of V initial, and this is the speed with which he threw the ball downwards, this is the data he would have found. Go ahead and look at this data, see if you can find the pattern and find what term belongs here. What term should be added to ½gt² to accurately describe the position x, which I've called d up here. Now, when you're entering your answer, to enter a number like V₀ you can just type the letter V followed by a 0, and the grader will know what you mean. Now, don't get discouraged if you can't figure this one out. It's a very tricky problem. But think about it, and if you do get it right, that's really impressive.
Well, let's see. Let's compare this entry to this entry. Every thing's the same except for the initial velocity. What's happened? Well, the object has gone 10 further. What about this entry to that entry? Well, again, every is the same except for the initial velocity, and the object's gone 20 m further. For this one and this one, the object's gone 30 m further. If we compare this entry to this entry, we've a 20 m increase, 40, and 60. Well, what do you know? In fact, this is always going to be the case. The distance an object moves when it has some initial velocity is equal to V₀ times the time plus ½gt².
This is our first serious equation of the course—Δx = V₀t + ½at². Now, I've introduced this Δ because this equation really describes how much and object position changes from here to there over a period of time, given an initial velocity and an acceleration. I wrote a here instead of g, because a is more general. In physics we like general equations. This is the acceleration of your object, which on earth in free fall does happen to be this number g, which is 10 m/s². On the moon, though, this number actually significantly smaller. So, this is one of the important kinematic equations. It's one of the equations that really describes how objects move. Now, there's a couple other kinematic equations that I want to talk about, and I'm actually going to do something that I'm a little reluctant to do. I'm going to give them to you. The next one I want to talk about is this one— velocity equals initial velocity + acceleration times the amount of time the object's been moving. Okay, fine. Why am I reluctant to give this to you? I'm reluctant because I already did. When we defined acceleration, we defined it as a change in velocity over some elapsed time, some change in time. In fact, that's exactly what this equation says, though it's hidden a bit. This velocity is actually what we can call some final velocity, after some time has elapsed. If we just rearrange this a little bit—like this—and then we can even divide by t if we'd like, we have this definition of acceleration. It's the change in velocity—final minus initial—over the elapsed time t. The last equation I want to talk about is this one. What does this one say? I says that there's some relationship between final velocity and initial velocity, acceleration and distance traveled. Deriving this one is a big trickier, but it's still something that I think you can do. I'm not going to do the derivation here, but I'd love to see some discussion about this in the forums. Now, let me say this is often the place where students get a bit scared. I've had many students tell me that equations are frightening, but the fact is they aren't. They're inanimate objects. They're not even objects. They're an inanimate idea. In fact, they're a tool. They're a tool for us. When you see a tool that you don't know how to use, you have two choices for how to respond. You can walk away, discouraged because you don't know what to do with it. Or you can imagine all the amazing things this tool will let you do, sit down, and learn to use it. I hope that's the option you choose. Let's sit down and learn how to use these tools.
We have three equations here that also supposedly describe the same thing--motion. But if they all describe the same thing, why do we need three of them? Well, let's think for a second. Motion so far seems to involve talking about these five quantities. How much our position changes, our initial velocity, our final velocity, our acceleration, and our time. If we go through each of these equations, we can see they each involve some combination of these variables. We can see that this equation involves all the variables except for the final velocity, this one everything except for change in position, and this one everything except for time. These are all going to be useful in different circumstances. Let's do some practice. Let's say I gave you this situation. Here we have Galileo again on the top of the leaning tower, which he may or may not have every done, and he's going to through a ball downwards at 5 m/s. Now, the height is 57 m. That's the change in position that the ball is going to have to undergo. Now, my question is out of these five variables, which three do we know? And we do know three, not only two.
Well, we know the initial velocity is 5 m/s. We know the height, and that corresponds to a change in position. If we're going to drop this ball 57 m, it's going to change position from here to here, a distance of 57 m. We know Δx and, though I didn't tell you this explicitly, we know the acceleration. Acceleration is just going to be that number we labeled g or 10 m/s².
Now, my question is how fast is this ball moving when it hits the ground. What is the final velocity? If I'm asking you about the final velocity, it means this is the fourth variable involved in the question. So, we have 1, 2, 3, 4 variables involved in this problem, and which four? Well, these four. Everything except for time. That looks like it corresponds to this equation, this tool. As every carpenter knows, having the right tool for the job makes this remarkably easy. Now we can just plug in. Can you use this new tool to plug in our known values and calculate V? Enter your answer in meters per second rounded to one decimal place.
Well, let's see. V² = V₀ or 5² + 2 times our acceleration, which is 10---g, times Δx, which is 57 meters. I get that V² is equal to 1165, and the unit there will actually be m²/s². When I take the square root, I find the velocity is 34.1 m/s. Let's review the strategy here--understand the problem, identify our variables, first we identify which were given to us and we have to be careful to recognize the acceleration was given to us, identify what's being asked. Do we know that these were the four variables in question? Find the corresponding equation--well, this one uses those four variables. And plug and chug. Not too bad. Let's do another practice question.
For this question we throw a ball straight up in the air at 20 m/s I want to know what's the highest point above the ground that the ball reaches. Now, this question is tricky in several ways. Let's first try to identify our knowns, what was given to us. Well, we threw it straight up with 20 m/s of velocity. That's got to be the initial velocity, because that's how fast it's going at the beginning of the problem. Now, since this is a problem that's taking place on earth, and the object is in free motion, I of course know the acceleration is 10 m/s². But then I'm sort of stuck. What is the highest point above the ground that the ball reaches? Well, highest point seems important, but I'm still a little confused and without a picture, I'm not going to make much progress on this question. So, let's draw a picture of what happens when you throw a ball straight in the air with some initial velocity. Well, it goes up, up, up, up, up. Eventually it stops and turns around and then comes back down. Ah, okay, so this highest point, I can see is some sort of change in position. The problem gives me the initial velocity, and I know the acceleration. I'll need one more piece of information if I'm going to solve for Δx. If you've been listening very carefully, I gave a hint. Which of these two variables do we actually know when the ball is at the top of its trajectory?
The answer is velocity. We said before the ball goes up, up, up, up and then stops for a brief instant. Stopping means at this point in the trajectory, the velocity is equal to 0. That's a trick we're going to have to use pretty frequently when we're solving these sorts of problems.
Now, going back to our equations, we see we're going to use the exact same one we used last time, because these four variables are the relevant ones, except this time we're solving for the height—Δx. Can you tell me what is the highest point above the ground the ball reaches? Enter your answer in meters to one decimal place.
Well, when we plug in everything we know, we know at the very top the velocity is 0, and 0² is just 0, equals the initial velocity squared. Well, 20² + 2aΔx, and Δx is our unknown. If we solve this we move the 20² over here--20² is 400, so we have -400, equals 20 * Δx, which means that Δx = - 20 m. Uh oh. I'm confused. What is this negative business?
This negative makes me think we may have made a mistake, so let's look at our variables. Well, we said the initial velocity was 20 meters per second. Yeah, that was absolutely true. We threw it up at 20 meters per second. We said the final velocity was 0. We can't really argue that. At the top of the trajectory, it does have to stop for a brief instant. And we said the acceleration was 10 and that's meters per second squared. When on earth isn't that always the acceleration? Well, here's where we have to be careful. When we said that the initial velocity was +20, we made a choice. We said, "Hey, in this problem, up is positive." Now, we were free to make that choice. We could've said, "Down was positive too." But once we make our choice, we're no longer free. Acceleration. Objects always accelerate downward. They accelerate towards the earth. They don't spring up off the earth. If we've chosen up is positive, we have to choose A as negative since it points downwards. And if we carry that out, we get a +20 here and so the maximum height is 20 meters. Remember whenever you make a calculation in physics, you are calculating something about the real world. If you're number doesn't make sense in the real world, you did something wrong. This is nice because it's a built-in sanity check. We can always check to see that our number isn't totally crazy, and we know if it is totally crazy we made a mistake. If it's not, we still may have made a mistake but we feel more confident with a better answer. One more question before we move on. And this time you throw a ball again straight up at 20 meters per second. But now I want to know how long does it take to reach the highest point. Enter your answer here.
Well, let's see. We know initial velocity. We know acceleration since we're in the earth. We can use the same trick to know the velocity if the top of the trajectory is zero and we're looking for the time--okay. These were variables correspond to this equation V = V at plus 80 plugging in my velocity at the top is 0 minus your velocity is 20 plus my acceleration which is - 10 times time. If I solve this, I find that time equals 2 seconds. That makes tons of sense. We know the acceleration due to gravity is 10 meters per second per second. If I start off with the velocity of 20 meters per second going up, I'm going to lose So it's going to take 2 seconds for this velocity to go down to 0 and that's the top of my trajectory.
Now let me say you've done amazing work if you've made it this far. You know everything there is to know about objects moving in a straight line. That is pretty cool. Now you should go ahead and take a breather maybe go outside and drop some balls off of leaning buildings or go to the forums and ask any of the questions you might have. I said amazing questions, but really I should have said any because in the world of questions every question is an amazing question. There's plenty of people in the forums who would love to help you out and once you're done there, you can come back and learn about two dimensional motion where you'll finally know how to kick a soccer ball to get it past the goalkeeper, where to aim your arrow so it hits the bulls eye and any other question that involves motion in two dimensions.
Welcome back; now, it’s time to talk about 2-dimensional motion and this is where things get really interesting. So far, we’ve only talked about 1-dimensional motion and that motion in a straight line either purely up and down motion or purely left and right motion. But what happens when we have some combination? Well, then we get the motion with—this is the sort of motion that describes a cricket ball as it bounces towards the batsman or an arrow as it sails towards it’s target or a parachutist who has just jumped out of a plane. Now, before we talk about our modern day understanding of 2-dimensional motion and in fact Galileo’s understanding, let’s talk about what Aristotle thought of 2-dimensional motion. In Aristotle’s mind--and this is Aristotle over here of course-- objects, for example, a ball thrown into the sky moving in a straight line until they ran out of what he called impetus. At that point, after the fall straight down and land somewhere on the ground. Now, before I continue, why don’t you try this. Throw a small object like a coin or a pencil in front your field of view, and look at what you see. It’s actually really difficult to see what’s going on. In fact, it wasn’t until Galileo came around that the true path of an object in 2-dimensional motion was known. That path is what called a parabola. So, we know that 2-dimensional motion is parabolic. And so what? Parabolic is just a word. What does that mean? In fact, can we use this word to somehow describe 2-dimensional motion mathematically? In physics, prediction is power, and what I mean by prediction is, can we develop our understanding of motion in such a way that we can explain and predict mathematically? How an object will move before we actually throw that object, let's say? We can and that’s what we will do in the rest of this unit.
How did Galileo quantify motion in 2 dimensions? Well, he was a good scientist so he did an experiment. This experiment involved rolling balls off of a the table and controlling the speed with which they rolled off the table and measuring where the ball landed. Galileo controlled the speed of the ball by adjusting the height from which it rolled. If he wanted it to go faster, he will roll it from higher up. We'll talk later about how exactly he can know where to put the ball. Galileo controlled the speed of the ball by adjusting where in the plane he released it from. Higher up of course meant it would be going faster by the time it got to the edge of the table. Like in his previous experiments, he didn't know the units of his velocity, but he knew their relative sizes. For example, he knew he can make the velocity twice as big as it was originally or three or four times bigger and look what he found. Again these are unit list distances but look at this correspondence. When the ball had a velocity of 1 unit, represented here, the trajectory looks something like this. With a velocity of 2 units, the ball went twice as far. Likewise with 3 and 4 units of initial horizontal velocity. Galileo called this Vx. This x indicates an x direction. When we label axis, we usually called the horizontal direction x and the vertical y and so this x is just here to remind us that we're talking about a velocity in the horizontal direction. Okay, this is pretty interesting. What's the conclusion that we can draw from this data? Is it that the horizontal velocity is equal to the motion in the x direction? That horizontal velocity is somehow proportional to the motion in the x direction? Or that horizontal velocity and x direction motion are totally unrelated? What's the best answer?
Now, this was a really tricky question. You might have been tempted to choose this first answer. I can understand why--1, 1, 2, 2--seems pretty equal. But not quite, because this left column has units of speed or velocity, and this right has units of distance. It's like comparing apples to oranges. They can never be truly equal. A better thing to say is that they're proportional, and that there's some constant of proportionality which relates the two.
We've learned that the faster the ball is moving to the right, the farther it goes to the right. That's pretty intuitive. Good. This should be intuitive. Galileo didn't just measure Δx though. He could also measure the time it took these balls to hit the ground. What he found was truly remarkable. Give this data, I want you to tell me what's the best relationship between horizontal velocity and the time the ball spent in the air. Are they equal, proportional, or totally unrelated?
The answer is that they're totally unrelated. This might not be what you expected. If a ball is going faster, maybe you think it'll spend more time in the air or less time in the air. But the data says, no, and in physics we have to listen to the data. This is hinting at a deep and beautiful fundamental truth of 2-dimensional motion.
The truth is that motion has two components--it has a horizontal and a vertical component-- and those two components are completely independent. What do I mean by independent? I mean that we can break a two dimensional motion problem into two separate one dimensional motion problem and this is an amazing tool. Then we go on and shown how in 1 dimension, if we're for example dropping a ball, at the instant when we dropped the ball, well it has no velocity not quite yet. One second later, it's picked up some velocity. It picks up that much again in the 2nd second and again it picks up the same amount in the 3rd, and we can represent that pictorially with pictures like these. Now what if instead of dropping the ball, we throw it to the side. What kind of errors are we going to draw now? The instance the ball was released, it only has horizontal velocity--that's from you throwing it. I said before that these components are completely independent so that means after 1 second this horizontal component--well, it doesn't change. It keeps going to the right and just like in the 1 dimensional problem before it picks some speed going downwards. In fact, it picks up the same amount of vertical speed that it would have had we just dropped the ball. After 2 seconds, horizontal component does not change but the vertical looks twice as long and after 3 seconds the vertical grows again and the horizontal stays the same. Now this is sort of a strange picture here. Is this ball moving to the right or is it moving down or is it moving in some other direction? Which of these errors best describes the direction of the balls' motion at this point in time? And it's a guess so don't worry if you're wrong.
The answer is this one, the ball is moving down to the right and just think if we were throwing a real ball up a tower, its path would look something like this. In any instant we can approximate the direction it's moving, a straight line like this and yeah sure enough it goes down to the right. Remember, whenever you ask a qualitative answer like this one in physics, well, physics describes reality. You should never provide an answer that disagrees with what you think would actually happen. Okay, so how do we get this answer. With this blue arrows are what we call components of the balls velocity. We've already mentioned the horizontal component and this one we'll call the vertical component. Those two together create the balls resulting velocity but they have to be added appropriately. The way we do that is we imagine sliding this horizontal vector down, so its tail matches with this guy's tip and we can also slide this vertical one over here. Now we've created this nice rectangular shape where those meet we draw our line and we have our resulting velocity. This is going to come in handy when we wanted to do some trigonometry later, and just to give you a hint we can already see the right triangles forming. Now this has been talked about in a very abstract sort of way. Let's actually solve a problem so we know what's going on.
Here's a general problem which is a classic 2-dimensional motion problem. Before we get into it though, let's remind ourselves to the equations we learned in 1-dimensional motion. Well, these equations relate a few variables, Δx, initial velocity, time, acceleration, and velocity which we called some final velocity. I'm going to write those over here. To fully understand an object's motion in 1 dimension, we like to somehow decide what the values were for all of these variables. On 2-dimensional motion, I said which is solving two of these one dimensional problems. So, we need to make some copies. Δx is going to partner with Δy. V0 I'm going to call V0x like I did over here to distinguish between whether the velocity is horizontal or vertical--its prime will be V0y. There's going to be two accelerations, one in the x direction and one in the y, likewise two final velocities. Interestingly enough, however, there is only going to be one time. There is no x and y version of time because unlike position, velocity, and acceleration, well, there is no easy way to assign a direction to time. In physics, we call time a scaler and these other quantities are vectors. You don't have to worry about the vocabulary for now. If we want to know the trajectory of this ball as it travels through the air, our goal is to get numerical answers for each these quantities. Some, we can already fill in. V0x I know be 10 meters per second. This is the number that I just gave you at the start of the problem. What about Δy? Can you tell me what delta y would be in meters? Well, that's the vertical distance the ball is going to move over the course of it's trajectory and that's exactly equal to 50 meters. What about this acceleration of 10 meters per second squared? Is that a horizontal or a vertical acceleration?
Well, it seems like we've used all the numbers that were given to us, the 10, the 10, and the 50. Now we're going to have to do some work. V0y, even though it wasn't given to use explicitly, is something we can actually answer right now without doing any math. What is V0y?
It's 0. The ball is only moving horizontally to begin with. This 0 indicates we're talking about the beginning of the problem, and the y means vertical direction. There's no vertical velocity at the beginning of the problem. So, V0y is 0 m/s.
Now, looking only at the numbers on the right side, because this is what we're familiar with from 1-dimensional motion, could you tell me what the time it will take for this ball to hit the ground is?
Well, we're going to need to use one of our equations. We're going to need to use the one that involves all of the variables except for final velocity. And that would be this equation. Okay, using only numbers from the right column, Δx, well that's actually Δy, 50 = V0 * t, but that's 0 * t. That's just 0. Plus ½, I can use my 10 for the a, times t². Solving this, we get a result of 3.2 seconds.
Now, what's the final y component of velocity--whatever that means? We'll talk more about what that means at the end of this problem. For now, let's just calculate what this Vy will be. Enter your answer here.
Well, we're just looking at these numbers, and the easiest equation to use for me would be this one—Vy, even though the y isn't here in the original equation, remember I'm just looking at this right-hand side, equals the initial velocity in the y direction, which was 0, plus the acceleration, which is 10. Well, Vy just equals the initial velocity in the y direction, 0, plus 10 times 3.2--acceleration times time. Giving that out gives us 32 m/s.
Look how much progress we've made. We've already filled up half of these values. We're learning a lot about the motion of this object. In fact, we already talked about how the time over here is the same over here. There's only one time in the problem. I can fill that in here, and I've already told you that acceleration is only straight down. If I drop an object, it doesn't spontaneously drift to the left or right. It goes straight down, meaning the acceleration in the x direction is 0. Now, can you, using only the numbers in this left-hand column and these equations fill in these remaining two entries? Enter your answers here and here.
Let's use this equation. Vx = 10, which is the initial x velocity, plus 0, which is the acceleration, time 3.2. Well, that goes to zero, and so we just have an x velocity of 10. Note that the initial and final x velocities were exactly the same. That is not a coincidence. That's due to the fact that there's no acceleration in the x direction. Now to find Δx I use this equation. The initial velocity is 10 times the time, which is 3.2, plus ½at². Note, again, a is 0, so this goes to 0. The object will land 32 m away from the wall. This is an amazing ability that you now have.
Now it's your turn. I set up a very similar problem to the last one. If you need help solving this one, you should go back and see how we solved the last one. But in this problem, we have, let's call him Archimedes, standing on the bluffs of Syracuse. He's firing his bow straight out into horizontally into the ocean to see how far a ship could be before he could reach it. His bow fires his arrow at 50 m/s, and the height of the bluff is 100 m. Can you fill out all of these fields here? If you can get this right, I will be very, very impressed.
Well, let's fill out what was given. We know the initial x velocity is 50. X acceleration is 0 as it almost always is. And the y acceleration is 10. Δy is just the height the object is going to move vertically--100 m. The initial velocity in the y direction is 0, because it's only moving horizontally. It's not moving vertically at all. We can use this equation to plug in these variables, and when we do that we get a value for t of 7.1 seconds, which is also the time over here. Again, making our appropriate substitutions, using this equation we find a final y velocity of 71 m/s. What about the x? Well, If I know the initial x velocity and there's no acceleration-- that is, the x velocity isn't changing-- the final x velocity has to be exactly the same--50 m/s. Using this equation and plugging in the appropriate values, I find a Δx of 355 m. If you got this right on your first try, I am very impressed. If you didn't, go back and try it again. If you're still confused, go to the forums and I'll be there along with a bunch of your classmates to help you out.
Now, this 355 m is pretty good. That's far for a bow and arrow. But my question is how can we shoot farther? If we want to shoot our arrow even farther, what would we do? Well, maybe it's the case that we just can't shoot any farther. Maybe we should aim straight up into the sky or maybe we should just aim slightly higher, so we shoot at some angle. What do you think is the best answer?
Well, the best answer is that we should aim slightly higher. That seems pretty intuitive. If I'm trying to throw a ball to my friend who's far away, I don't throw horizontally. I throw slightly into the air. Now, the question is how much higher do we aim? As we always want to do in physics, let's quantify.
Let's consider a problem where I throw the ball with an initial velocity of 50 m/s. But this time I don't say V₀x or V₀y, because this is not purely horizontal or purely vertical. It's at an angle. Now, we know if we can break this problem into two 1-dimensional motion problems we're all set. So, that's should be our end goal. Let's pause for a second. Not literally. Don't hit the pause button. But let's think back to when we first started talking about horizontal and vertical velocity. Remember how we talked about a ball with vertical and horizontal velocity components? Well, these components were just a way of thinking about the true velocity, which was off at some angle. In that case, we went from components to the true velocity. Now, we're going to go the other way. What do I mean by that? I mean, we're going to make a right triangle. How do we do it? Well, here's our angled side. Here's our hypotenuse. Now, I've broken this hypotenuse down into what I call components-- a horizontal and a vertical initial velocity component. Now, if I call this angle α, and I tell you that α is equal to about 53 degrees, can you tell me what V₀y equals? You might have to think back to the trigonometry from unit 1. Enter your answer here and round to the nearest whole number.
Well, here we have an angle, and we have an opposite side and a hypotenuse. Opposite and hypotenuse--well, that's the sine relationship. In fact, I can write that sin α is equal to V0y over 50. It's the opposite over the hypotenuse. Solving this means that V0y is equal to 50 times the sin α, which is 53 degrees. If I type this into a calculator or a search engine, I get 39.9 or rounding 40 m/s. I can do some similar trigonometry using the cosine to find that V0x is 30 m/s. Now, look what we've done. We've broken one very difficult problem into two very solvable problems. Very cool.
Well, let's solve the problem. There's a few fields I think you can fill in right now. Can you tell me the horizontal and vertical accelerations as well as the horizontal and vertical initial velocities?
As always, horizontal acceleration is 0, vertical is 10 m/s². Horizontal velocity, what we just calculated, is 30, and the vertical is 40 m/s.
Now I'm actually going to move this picture out of the way. It's actually become distracting with the person and all this unnecessary information. I'm just going to draw this diagram, which boils out the essential information. One thing I'd like to point out before we solve this problem—Δy. In reality, the ball starts at ground level, goes up to some maximum height and then ends at ground level. Since it starts and ends at the same altitude, well, Δy is really equal to 0. So, maybe I shouldn't call it Δy. Instead, let's solve for the maximum height the ball reaches. Just like I did before, I'm going to approach this problem by looking at the vertical component first. Now, actually, this seems like it could be pretty difficult. I've only got two pieces of information, but that's not entirely true. So, let's exploit the one other fact we know about 1-dimensional motion. We know that when an object reaches the absolute peak of its trajectory it's vertical velocity is 0—a.k.a. at this point vy = 0. It's also true that an object spends just as much time going up in its trajectory as it does going down. That's not something I'm going to prove for you here, but it's something you should definitely talk about in the forums. What's also true is that when a ball starts and ends from the same altitude, same height, it takes the exact same amount of time to go up as it does to go down. You could say t-up equals t-down. Okay, now we're getting somewhere. Let's use this equation. Well, I know at the highest point the balls vertical velocity is 0. The total velocity isn't 0, because it's still moving towards the right, but it's not moving at all vertically. That equals the initial vertical velocity plus acceleration times time, and this is where we have to be careful. Acceleration here is -10. Once I choose the initial velocity upwards of 40, acceleration is downwards has to be negative. Times time. If I solve this I find that the time is equal to 4 seconds. Now, can you tell me what I should put here if this time represents the total travel time of the ball going up and then back down? I'll give you a hint. The answer is not 4.
The answer is 8, because it took 4 seconds for the ball to go up, and it's going to take another 4 to go down. That's also the time over here—there's only one time.
Well, we can use this equation, and since acceleration in the x direction is 0, this term goes to 0, and we just get 30 * 8, which is 240 m.
Oh, no! I've realized we forgot to calculate height. Can you go back for me and calculate the maximum height this ball reaches? Be careful. You may not want to use 8 seconds for your time.
Well, the height is just a change in vertical position. It's a Δy. We can use this equation. In fact, I can say that height equals the initial velocity times the time, and I've used 4 seconds, because it only takes half the time to get to the top, and the other half is spent going down, plus ½ times the acceleration times the time². I get a result of 80 m.
Now I can actually use this equation again to solve for Vy. That would be the final velocity after 8 seconds just before it hits the ground. Can you solve for Vy? Be very careful with your sign. Is this going to be a positive or a negative velocity? Enter your answer here.
Let's see. Vy equals—well, the initial velocity was 40—plus -10 times 8 seconds.
Let's go in and fill out the left-hand column. What's the final velocity in the x direction going to be?
Well, the initial x velocity is 30, and there's no x acceleration. We could use another equation, or we could just say if there's no acceleration velocity isn't changing.