Projectile motion (part 2)

In the last video, I dropped myself or a penny from the top of a cliff. We started off at 0 velocity, obviously because it was stationary, and at the bottom, it was 100 meters per second. We used that to figure out how high the cliff was, and we figured out that the cliff was 500 meters high. What I want to do now is let's do that same problem, but let's do in a general form, and see if we can figure out a general formula for a problem like that. Let's say that you have the same thing, and let's say the initial velocity-- you're given the initial velocity, you're given the final velocity, you're given the acceleration, and you want to figure out the distance. This is what you're given, and you want to figure out the distance. Doing it the exact same way we did in that last presentation, but now we're now [INAUDIBLE] formulas, we know that the change in distance is equal to the average velocity times-- we could actually say the change in time, but I'll just say it with time, because we always assume that we start with time equals 0-- times time. We know that the average velocity is the final velocity plus the initial velocity divided by 2, so that's the average velocity-- let me highlight-- this is the same thing as this, and then that times time. What's the time? You could figure out the time by saying, we know how fast we're accelerating, and we know the initial and final velocity, so we can figure out how long we had to accelerate that fast to get that change in velocity. Another way of saying that, or probably a simpler way of saying that, is change in velocity, which is the same thing as the final velocity minus the initial velocity is equal to acceleration times time. If you want to solve for time, you could say the time-- if I just divide both sides of this equation by a-- is equal to vf minus vi divided by a. We could take that and substitute that into this equation, and remember-- this is all change in distance. We say change in distance is equal to-- let me write this term in yellow-- vf plus vi over 2. Let me write this term in green. That's times vf minus vi over a. Then if we do a little multiplying of expressions on the top-- you might have recognized this-- this would be vf squared minus vi squared, and then we multiply the denominators over 2a. So the change in distance is equal to vf squared minus vi squared over 2a. That's exciting-- let me write that over again. The change in distance is equal to vf squared minus vi squared divided by 2 times acceleration. We could play around with this a little bit, and if we assume that we started distance is equal to 0, we could write d here, and that might simplify things. If we multiply both sides by 2a, we get-- and I'm just going to switch this to distance, if we assume that we always start at distances equal to 0. di, or initial distance, is always at point 0. We could right 2ad-- I'm just multiplying both sides by 2a-- is equal to vf squared minus vi squared, or you could write it as vf squared is equal to vi squared plus 2ad. I don't know what your physics teacher might show you or written in your physics book, but of these variations will show up in your physics book. The reason why I wanted to show you that previous problem first is that I wanted to show you that you could actually figure out these problems without having to always memorize formulas and resort to the formula. With that said, it's probably not bad idea to memorize some form of this formula, although you should understand how it was derived, and when to apply it. Now that you have memorized it, or I showed you that maybe you don't have to memorize it, let's use this. Let's say I have the same cliff, and it has now turned purple. It was 500 meters high-- it's a 500 meter high cliff. This time, with the penny, instead of just dropping it straight down, I'm going to throw it straight up at positive 30 meters per second. The positive matters, because remember, we said negative is down, positive is up-- that's just the convention we use. Let's use this formula, or any version of this formula, to figure out what our final velocity was when we hit the bottom of the ground. This is probably the easiest formula to use, because it actually solves for final velocity. We can say the final velocity vf squared is equal to the initial velocity squared-- so what's our initial velocity? It's plus 30 meters per second, so it's 30 meters per second squared plus 2ad. So, 2a is the acceleration of gravity, which is minus 10, because it's going down, so it's 2a times minus 10-- I'm going to give up the units for a second, just so I don't run out of space-- 2 times minus 10, and what's the height? What's the change in distance? Actually, I should be correct about using change in distance, because it matters for this problem. In this case, the final distance is equal to minus 500, and the initial distance is equal to 0. The change in distance is minus 500. So what does this get us? We get vf squared is equal to 900, and the negatives cancel out-- 10 times 500 is 5,000, and 5,000 times 2 is 10,000. So vf squared is equal to 10,900. So the final velocity is equal to the square root of 10,900. What is that? Let me bring over my trusty Windows-provided default calculator. It's 10,900, and the square root. It's about 104 meters per second, so my final velocity is approximately-- that squiggly equals is approximately-- 104 meters per second. That's interesting. If I just dropped something-- if I just drop it straight from the top-- we figured out in the last problem that at the end, I'm at 100 meters per second. But this time, if I throw it straight up at 30 meters per second, when the penny hits the ground, it's actually going even faster. You might want to think about why that is, and you might realize it. When I throw it up, the highest point of the penny-- if I throw it up at 30 meters per second, the highest point of the penny is going to be higher than 500 meters-- is going to make some positive distance first, and then it's going to come down, so it's going to have even more time to accelerate. I think that makes some intuitive sense to you. That's all the time I have now, and in the next presentation, maybe I'll use this formula to solve a couple of other types of problems.