 Hey everybody, welcome to Tudor Terrific. This is the final video, Lesson 5, in my second unit in my physics course on one-dimensional motion. In this final video for this chapter, we're going to look at falling objects. So we're going to bring acceleration due to gravity specifically into our discussion of one-dimensional kinematics. We're still going to be looking at doing problems with the big three. So let's get started. All right, Galileo Galilei, right in the height of the Renaissance and the early Brogue period. He was an amazing scientist. We give him credit to the telescope and many other things in physics. And one in particular pertains to us today. Galileo's famous experiment involved dropping two different sized objects with different masses off the leading tower of Pisa in Pisa, Italy. Now, most people thought that the larger object with more mass would hit the ground first. Galileo theorized that that was not true, that both objects regardless of mass would hit the ground at the same time. What do you guys think happened? Well, to test this theory, we have to ignore air resistance. Objects with a larger surface area, when they contact a larger area of air, a cross-sectional area of air, will retrieve more drag, more air resistance. So what was done in the Apollo landings was, I think this was the final Apollo mission. They took a video, the astronauts of a famous experiment by Galileo, this famous experiment, but done with a hammer and a feather. On the surface of the moon. Now, the surface of the moon is a little bit different than the surface of the earth because the air resistance in the atmosphere is almost negligent on the moon. And so both these objects would fall like in a vacuum. Some physics teachers like to do this experiment with an actual vacuum and you can see quite an astounding result. So on earth, in our atmosphere, that hammer he's got in one hand would fall a lot faster than a feather. He got in the other hand because the feather would react differently to the air resistance. It would fall much slower, it would fall in a wave-like fashion. But watch what happens when he lets both of them go on the surface of the moon. They both hit the surface of the moon from a height of about four feet at the exact same time. Now that's what would happen to objects on the surface of the earth in the absence of air resistance. And Galileo was able to actually show that things with, I bet the ball with smaller surface area actually felt just a tad bit faster than the larger ball. And this was quite an astounding result. And it showed the world that all objects fall with the same acceleration. So, and that's acceleration due to gravity. Now I'm not going to talk too much about gravity except for this acceleration. Okay, let's say we have a little pepper here, held by one of my former students. And we have a vector. I'm going to name that an acceleration vector, so that's why it's yellow. And I'm going to call it G. That's the arrow I give. And the name for the arrow I give to acceleration due to gravity. That pepper, if she lets it go, will fall to the earth, to this lake with that acceleration. We have a giant rock. Let's say we let it go at the exact same time. In the absence of air resistance, it'll also fall with that same acceleration vector, G, okay? All objects fall at the same rate of acceleration if we ignore air resistance. This is something Galileo theorized and he is correct. Now, it depends actually on many other factors we're going to look at in later chapters. But near the earth's surface, we can approximate the value of G as a constant. That constant is this right here, 9.8 meters per second squared. 9.80 meters per second squared. That's its approximation, okay? So if you're near the earth's surface, you use that value for the acceleration due to gravity. Or in other words, G, that's what it's commonly called, okay? It will always point directly toward the center of the earth. Now, if you're not drawing the entire earth, just point it straight down, okay? Now, the most important part is that that air is the same size. That G vector is the same, 9.8 meters per second squared. If we ignore air resistance for any size object, an object with any mass. That's the truth in physics. So we're going to use this in relation to the big three. It'll be our acceleration vector for our problems today. Okay, so we have a problem falling from a tower. Suppose that a ball, like in Galileo's experiment, was dropped from a tower 70 meters high. How far will the ball have fallen after a time of one second, two seconds, and three seconds? If it's falling, we assume with gravity's acceleration. Okay, like normal, we will look at our given and wanted tables. But, but, we must first do something we have not done yet. We must decide a positive and negative direction for our problem, okay? Because some quantities, when it comes to free fall and everything you're going to do after this, can be negative depending on our definition, okay? Because we're dropping something from a height 70 meters up, and it's falling the entire time, I'm going to create a positive direction that makes most of my variables positive. Because I'd rather work with positive values than negative values. It's just easier. So I'm going to make positive downward, and that's why I draw this arrow like this, okay? I draw it downward with a plus sign after the tip, so I mean that downward is positive. Okay, now let's look back at what we're given. We're given an initial velocity, that's what it means. When dropped, it means the initial velocity is zero. We're given the initial position we could say is 70 meters. And the final position is not clear, but we get a bunch of different times. And we know, since it's being dropped near the surface of the earth, that the acceleration is 9.8 meters per second squared. So if I treat positive downward, I'm going to have to change some of these initial values, okay, to make the problem easier. But what I want is how far the ball will have fallen after these times. So that would be the final position, okay, with t at these different values, okay? That's why I have them here. But let's change some things to make it easier. We can set a reference point wherever we want in these problems. So I'm going to choose to set the reference point at the top of the tower. Why would you do that? Why would you say that's zero? Well, you've got to think differently than measuring the height of a tower. You've got to think more mathematically. You can set the reference point where you want. If positive is downward, then the position will grow as the object falls, okay? So it starts up here at zero, and it goes down, down, down. It hits the ground. It'll be at positive 70 meters in this situation. It's perfectly fine to do. Also, I'm going to set my acceleration to positive 9.8 meters per second, because that acceleration vector in this situation points down like my positive direction is. So the acceleration, since it points in that same direction, is now positive. Okay. In order to get the final position, given all of these things, the initial velocity, the initial position, the acceleration, and the time, you can see that the second of our big three kinematic equations is the one to use, okay? The variable we want is already isolated, and all of these on the right side we know and can plug in, okay? Right off the bat, we can make the situation simpler because our initial velocity is zero, so the middle term is zero, and x naught is zero. So we could put a zero there as well. It's a very simple one-half a t squared. And one other thing that we like to do, and I'm doing right now, is I'm switching all my x's into y's. The reason I do that is because we are thinking vertically, and since we worked so much in math in the Cartesian coordinate plane, you might want to think in terms of y for vertical free fall or falling object situations. So that's why I've made that change, okay? So the middle term is zero, and the first term is zero, so all we have is that the final position is equal to one-half a t squared. Don't mind the ones. Those ones aren't necessary, except they relate back to t one here. So we're going to plug in that first time one second in for t, and we will get one-half times 9.8 times one squared, which gives us 4.9 meters. That's how far it falls in one second from its 70 meter height. Now, the second position will be after two seconds. So let's see if it falls another 4.9 meters or it falls more than that. It's accelerating, remember. It's not falling at a constant velocity. That's exactly what happens when you plug in two. You get one-half times 9.8 times two squared, which is four. So this is four times what the previous answer was, 19.6 meters. It's falling at a faster and faster rate as time continues. Let's see if that pattern continues. Let's plug in three seconds, okay? So one-half times 9.8 times three squared, which is nine. So this would be nine times the original, 44.1 meters, okay? So as the ball falls or the object falls, the more time passes, the more difference in position, more displacement is accrued and not at a linear rate, an accelerating rate, quadratically speaking, because t is squared. Okay, let's look at another problem. This time we're not going to drop a ball from the top of a tower. We're going to throw a ball upward into the air with an initial velocity. And that's what's given, the initial velocity of 15 meters per second upward. We're going to calculate first how high it goes, and then we're going to calculate how long the ball is in the air before it comes back to the person's hand, assuming it does. And all these problems, I have not talked about air resistance factored in its mathematical changes into the problems. So we assume that air resistance will be ignored in these problems. So again, the given and wanted lists are needed, but we must first choose a direction that's positive. Since I am initially throwing the ball upward, I'm going to make my initial velocity direction as the positive direction. So now positive is upwards, okay? Let's write down what we know. We know the initial velocity is positive upwards, 15 meters per second. We can assume that the initial position is zero. Why can we do that? Because we could set the reference position wherever we want. So I'm going to set it at the initial position of the ball in my hand. Now, for part A, we want to calculate how high it goes. Now, if you look at a trajectory of this, it would be just the first portion of the ball's trajectory. At the top, where it's at its maximum height, it's not moving because it's turning around to fall. So at its maximum height, we can say the velocity is zero meters per second. That's a pattern you will continue to see throughout this course. Now, the acceleration vector, which has a magnitude of 9.8 meters per second squared, points down. It points down, points the opposite of my positive direction that I've just defined earlier. So the acceleration will then be negative 9.8 meters per second squared. Now, what do I want? I want the maximum height. So I want the final position with these givens, okay? The first section of the ball's entire trip. Now, again, I'm going to also write down what else I do not know to help me decide which equations to use. I have not given any time information for this problem. So I want that final position, and I'm given this information. Without time, you know to go straight to the time independent equation, okay? If you can, and here we have all the other quantities, all both velocities, the acceleration, and one of the positions. We can find x final here, which are the same as x here, easily using the third equation. But I'm going to, again, so that people understand, I'm going to replace all my x's with y's, so that people can see that we're dealing vertically. There's no horizontal stuff going on, only vertical. Also, the initial position is zero, so I can place a zero there to simplify the situation. Now, when I solve for y, I'm going to subtract v0 squared to the other side, and then I will divide by 2a, and y minus zero, or just y, will be by itself. So for part a, this is the setup. v squared final minus v not squared over 2a. Now, the final v we already know is zero, so that's plugged in there, minus the initial v, 15 meters per second. Don't forget to square the velocities, guys, in the third equation. Then we divide by 2 times the acceleration. Yes, you must include this negative here. You must do it. It is negative for a reason. If we did not, we'd get a negative answer for our position, which would make no sense. It would be underground to that ball, and that would be silly. So the ball will reach 11.5 meters with the proper sig figs looked at here after it reaches its maximum height. So when it's not moving at the top to turn around, it's 11.5 meters higher than when I threw it. Okay, part b, how long is the ball in the air before it comes back to its hand? Well, we need to adjust some things because the final position is now back at the start, okay? So the final position is now known, zero meters. What's not known is how fast it's going at the bottom. Nothing like that is given. Some of you astute physics people will already determine that it has something to do with this 15 meters per second. We're going to look at that in the next slide, but not yet. Let's just assume we don't know the final velocity right when it comes back to its hand. We are asked for the time for the entire trip now. So it's the entire trip up and down. Okay, the acceleration is still negative 9.8 meters per second squared. I see no reason to adjust my positive direction even though it's only spending half the time going up. That's fine. The initial velocity is still the same. Now, we want time. So using the third equation again would mean nothing to us. It would not help us. We will have to use the second equation. Why can't we use the first one? Gosh, I'm so sick of using these more complicated ones. Well, you can't use the first one because you don't know the final velocity. You have two unknowns in the first equation, T and VF. I can't use an equation if I have two unknowns. But in the second equation, the only unknown is T. I know these other things, okay? I'm going to change them so I can prove that I know them. The initial position y naught is zero. The final position y is zero as well. So this simplifies quite a bit to just two terms. But you realize that we're solving for T and this is quadratic in terms of T. However, the constant term is not there. Thankfully, thankfully, if we were just solving this quadratic equation for T like we would in Algebra 2. And you can see that both terms have a T in them so they'll be factored out. So I can factor out one of the T's and then I have this equation. Now, by the zero product property, I can set the T out here equal to zero separate from this parentheses here and I can set that entire parentheses by itself equal to zero. When you do that, you get the following results. From this tiny factor here, T equals zero. I just get T equals zero. And from this more complicated one, I would set this parentheses equal to zero and move the 15 meters per second to the other side. It would be negative and then I divide that by negative 4.9 meters per second squared and the negatives would dissolve. So you get this positive fraction and you get this time. Now, the first time, meaning zero seconds, is meaningless. It comes from the fact that this is a parabola. The shape of the position curve is a parabola and so it spends, it has two instances when it's at zero and that includes the start. So we ignore that for now because of course it's at its hand when it starts. The other time is when it comes all the way back down to his hand and that's 3.06 seconds. So that's the quantity that's reasonable and makes sense according to the problem because it asks before it comes, how long is the ball in the air before it comes back to his hand? So the moment it hits his hand is right here. So that's how much time it spends in the air. Okay. Now, those of you who were interested in the symmetry of this problem will like this. We're going to consider again the same situation I provided a picture here for you. We're going to make some more calculations, okay? First, I want us to calculate how much time it takes for the ball to reach the maximum height, all right? We know the entire trip's length of time but how much time does it take for the ball to just get to the top? Some of you might say half. Let's see if that's correct, okay? Yes, that is correct, okay? Why does that make sense? Because the acceleration is constant, okay? Because the acceleration is constant that means that the situation itself is symmetric, okay? The entire trajectory can be cut into two halves and they look identical except one is in reverse if you were to play it back and that's the time it takes to fall. So I asked my students when I'm in class what this is and they all say it's half. They notice the symmetry, okay? So the symmetry is there. Exactly half the time it takes to get to the top as it does to go all the way back down to the bottom. When we have gravity is our acceleration vector, a constant acceleration downwards. So 1.53 seconds is our time for reaching maximum height. So what you can do always is you can find the time for the whole trip and divide it by two if that's something that works easier than finding the time using these equations for half the trip. All right, now part B asks us the velocity of the ball when it returns to the thrower's hand, okay? What is the velocity of the ball when it hits, right before it hits his hand? Is it related at all to the 15 meters per second initial velocity? Okay? Now we are going to have to modify our given and wanted tables from the previous problem or just make them again so that the final values are for when the ball hits this person's hand again, okay? So here's our tables. Given we have the initial velocity and the negative 9.8 meters per second squared because positive is still upward for me, the time for the entire trip is 3.06 seconds. We're looking at the entire trip for part B, okay? Velocity of the ball when it returns to the thrower's hand. The initial and final positions. Now again, you can use x's or y's. I'm going to use x's this time. Zero, okay? The final and initial positions here are the same and I set them as the reference of zero. So I want that final velocity v. So let's continue. The top equation. Look at all that you know. You know so much. All you don't know is v. Since I have v naught, a and t, I can use the first equation to find that final v. It's the simplest way to do it. Of course, you could use the third equation as well, but why? It's more complicated. Use the first one. There's no squares. It's just a simpler equation. So, we don't have to solve for anything. We could just plug and chug v naught plus at in there right away, okay? 15 meters per second for v naught. A is negative 9.8 meters per second squared, hence the minus sign, times the total time of 3.06 seconds. Boom, look at that. The velocity right before it hits this hand is negative 15 meters per second. So it has the same magnitude of the initial velocity, but the opposite direction, like we had hoped for. That is the true symmetry of these type of free fall problems, okay? We're going to look more at that symmetry later on, but for now, I'm going to leave it at that. The time it takes to get to the top is half the time for the entire trip, and if the initial and final positions are the same, the velocity at the bottom will be the same magnitude, but opposite direction is the initial velocity. All right, guys, keep this stuff in mind when you're doing problems that involve free fall. You have to choose a positive direction, making sure that the vectors that point in the opposite direction of the one you chose as positive get negative values. We just saw that in the previous problems. You can also choose a reference point for your positions that make the calculations the easiest. We did that in all these problems. Sometimes we chose where the ball was leaving the person's hand at zero meters, and in the earlier problem, we chose the top of the tower as being zero meters because we chose positive downwards so that our position would increase as it fell. And also remember that the magnitude of your acceleration g is 9.8 meters per second squared. All right, guys, there is the end of Unit 2 on one-dimensional kinematics. Thank you so much for watching. The next unit on two-dimensional kinematics is coming soon. For now, this is Falconator signing out.