>> Okay, let's say you were going to toss a ball in the air to your friend that's hanging out the window of the dorm. And you want the ball to just be able to reach the friend such that they can reach out and grab it. It doesn't have any residual speed when it gets to them. How fast do you need to toss it? What is this Vi that you need to toss it at? And, obviously, we're going to have to know a few things. We're going to have to know the height that the person is at. We're also going to need to know what is the initial height that you launch it from, okay? So, how do we do this? Well, let's write down some givens, okay? Let's say that the mass of the ball is 200 grams. Let's say that our initial height is, basically, the height of a person that's about 6 feet which is about 2 meters. And let's say the stories are, maybe we have, let's see it's about 10 feet per story. So, let's say it a 3 story building, okay? So, that would be 30 feet, which is roughly 10 meters, okay? And let's ask this question. Vi is equal to what? All right, I think we could attack this a couple ways. One is you could go back and use the kinematic equations, that would certainly do it. But there might be an easier way which is the topic of this next module which is conservation of energy. So, when it is launched, it has some initial energy. When it gets to the top, it has some final energy, okay? The initial energy is composed of 2 things, right? It has some kinetic energy, one-half mVi squared, but it also has some initial gravitational potential energy. Because, if we're setting our Y equals 0 to ground level, then it's starting from some height Yi and so we need to include that in the initial energy. Now, when it just gets to the top, and it's no longer moving, right, it just comes to rest as it reaches your friend, then it is all potential energy. There's no kinetic energy left. And so the right side of this equation just becomes mgh. And now we want to solve this equation for Vi. All right, that's not so bad, right? We have Ms everywhere, cross them out. We want to get Vi alone, so let's multiply everything by 2. If I do that, I get Vi squared plus 2gYi equals 2gh. And now I can just rearrange some things and take the square root and you can probably convince yourself fairly quickly that this is going to be 2 times g times h. And then, when I subtract this over to the other side, this becomes a minus Yi. And now, at this point, I can punch in all the numbers, right? So, let's try it. We've got 2 times 9.8 times h, we said was 10 meters. Yi was 2 meters. Let's make a little room and let's calculate what this is. So, this becomes 2 times 9.8 which is 19.6. I'm going to multiply that by 8, 19.6 is pretty close to 20, right. Twenty times 8 would be 160, but it's a little bit less. So, let's say 156. Okay, and that is meters per second. That sounds really fast, right? Why does it sound really fast, because we forgot something here. What did we forget? We forgot this little square root, right? This was Vi squared. So, when we write this down here, we need to include the square root. And, if I do the square root of 156, that's got to be a little bit bigger than 12. That's my guess. Let's plug it into the calculator and see what we get, 156 -- well, let's do it properly, 19.6 times 8 is 157, so our guess was pretty close. We take the square root of that, and we get 12.5. Okay, 12.5. I'll put it up here so everybody can see it. Vi equals 12.5 meters per second. And that's how fast you would have to toss it up to your friend. Now, let's ask a follow-up question. Let's say that, after your friend grabs it, they then let it go. They drop it from rest and it falls all the way down to the ground. And let's see how fast it is going when it hits the ground. Remember it left our hand at 12.5 meters per second. When it comes back down to our height, it should be going 12.5 meters per second. But, of course, it's got to go a little bit further to get all the way to the ground. And so it's going to be going a little bit faster than 12.5 seconds -- 12.5 meters per second. So, let's see if we can calculate it. In the first part of the problem, we started here, we ended there. But, in this part of the problem, we're going to start at the top, and we're going to end at the bottom. And we can, again, use conservation of energy, Ei equals E final. Ei is all potential energy, mgh. E final is all kinetic energy, it's at the ground. We can solve this now for Vf pretty easily, the Ms cancel out. Multiply by 2, we get 2gh equals Vf squared. And then I take the square root. So, Vf equals square root of 2gh. Two times 9.8 times our height, which we said was 10 meters, 9.8 times 2 is 19.6 times 10 is 196. So this is the square root of 196. And, if you just punch that into your calculator, you should get 14. Fourteen meters per second which is, indeed, faster than we had before. Remember, originally, we had 12.5 meters per second Remember, originally, we had 12.5 meters per second for how fast we tossed it up. So, it picked up a little extra speed on the way down.
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10. Conservation of Energy
Solving Projectile Motion Using Energy
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