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Projectile motion (part 4)

Solving for time when you are given the change in distance, acceleration, and initial velocity. Created by Sal Khan.

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Video transcript

We'll now use that equation we just derived to go back and solve-- or at least address-- that same problem we were doing before, so let's write that equation down again. Actually, let's write the problem down. Lets say I have the cliff again, and so my initial distance is 0, but it goes down 500 meters. I'm not going to redraw the cliff, because it takes a lot of space up on my limited chalkboard. We know that the change in distance is equal to minus 500 meters. I'm still going to use the example where I don't just drop the ball, or the penny, or whatever I'm throwing off the cliff, but I actually throw it straight up, so it's going to go up and slow down from gravity, and then it will go to 0 velocity, and start accelerating downwards. You could even say decelerating in the other direction. The initial velocity, vi, is equal to 30 meters per second, and of course, we know that the acceleration is equal to minus 10 meters per second squared-- it's because acceleration gravity is always pulling downwards, or towards the center of our planet. If we wanted to figure out the final velocity, we could have just used the formula, and we did this in the last video-- vf squared is equal to vi squared plus 2ad. Now what I want to do is use the formula that we learned in the very last video to figure out-- how long does it take to get to the bottom, and to hit the ground? Let's use that formula: we derived that the change in distance is equal to the initial velocity times time plus acceleration time squared over 2, and that's initial velocity. The change in distance is minus 500, and that's equal to the initial velocity-- that's positive, going upwards, 30 meters per second, 30t. I'm not going to write the units right now, because I'll run out of space, but you can redo it with the units, and see that the units do work out. When you square time, you have to square the time units, although we're solving for time. Then, plus acceleration, and acceleration is minus 10, and we're going to divided it by 2, so it's minus 5t squared. We have minus 500 is equal to 30t plus minus 5t, and we could just say minus 5t squared, and get rid of this plus. At first, you say, Sal-- there's a t, that's t to the first, and t to the second, how do I saw solve this? Hopefully, you've taken algebra two or algebra one, in some places, and you remember how to solve this. Otherwise, you're about to learn the quadratic equation, although I recommend you go back, and learn about factoring in the quadratic equation, which there are videos on that I've put on Youtube. I hope you watch those first if you don't remember. We can do this-- let's put these two right terms on the left hand side, and then we'll use the quadratic equation to solve, because I don't think this is easy to factor. We'll get 5t squared minus 30t minus 500 is equal to 0-- I just took these terms and put them on the left side. We could divide both sides by 5, just to simplify things, and so we get t squared minus 6t minus 100 is equal to 0. I could do that, because 0 divided by 5 is just five, so I just cleaned it up a little bit. Let's use the quadratic equation, and for those of us who need a refresher, I'll write it down. The roots of any quadratic-- in this case, it's t we're solving for-- t will equal negative b plus or minus the square root of b squared minus 4ac over 2a, where a is a coefficient on this term, b is the coefficient on this term, negative 6, and c is the constant, so minus 100. Let's just solve. We get t is equal to negative b-- so negative this term. This term is negative 6, so if we make it a negative, it becomes plus 6. It becomes 6 plus or minus the square root of b squared, so it's minus 6 squared, 36, minus 4 times a, and the coefficient on a is here, and that's just times 1. With 4ac, c is a constant term, minus 100, minus 4 times 1 times minus 100, and all of that is over 2a-- a is 1 agains, so all of that is over 2. That just equals 6 plus or minus the square root-- this is minus 4 times minus 100, and these become pluses, so it becomes 36 plus 400. So, 6 plus or minus 436 divided by 2. This is not a clean number, and if you type into a calculator, it's something on the order of about 20.9. We can just say approximately 21-- you might want to get the exact number, if you're actually doing this on a test, or trying to send something to Mars, but for our purposes, I think you'll get the point. I'll say it approximately now, because we're going to be a little off, but just to have clean numbers, this is approximately 21-- it's like 20.9. We'll say 6 plus or minus-- let me just write 20.9-- 20.9 over 2. Let me ask you a question: if I do 6 minus 20.9, what do I get? I get a negative number, and does a negative time make sense? No, it does not, and that means that somehow in the past-- I don't want to get philosophical-- the negative time in this context will not make sense. Really, we can just stick to the plus, because 6 minus 20 is negative, so there's only one time that will solve this in a meaningful way. Time is approximately equal to 6 plus 20.9, so that's 26.9 over 2, and that equals 13.45 seconds. That's interesting. I think if you remember way back, maybe four or five videos ago, when we first did this problem, we just dropped the penny straight from the height. Actually, in that problem, I gave you the time-- I said it took 10 seconds to hit the ground, and we worked backwards to figure out that the cliff was 500 meters high. Now, if you're here at the top of a 500 cliff, or building, and you drop something that has air resistance-- like a penny, that has very air resistance-- it would take 10 seconds to reach the ground, assuming all of our assumptions about gravity. But if you were to throw the penny straight up, off the edge of the cliff, at 30 meters per second right here, it's going to take 13.5-- roughly, 13 and 1/2 seconds-- to reach the ground. It takes a little bit longer, and that should make sense because-- I have time to draw a little picture. In the first case, I just took the penny, and its motion just went straight down. In the second case, I took the penny-- it first went up, and then it went down. It had all the time when it went up, and then it went down a longer distance, so it makes sense that this time-- this was 10 seconds, while this time was 13.45 five seconds. You can kind of say that it took-- well, you actually can't say that. I don't want to get too involved, but I hopefully this make sense to you. If you have a smaller number here, you should have gone and checked your work, because why would it take less time when I throw the object straight up? Hopefully, that gave you a little bit more intuition, and you really do have in your arsenal now all of the equations-- and hopefully, the intuition you need-- to solve basic projectile problems. I'll now probably do a couple more videos where I just do a bunch of problems, just to really drive the points home. Then, I'll expand these problems to two dimensions and angles. Before we get there, you might want to refresh your trigonometry. I'll see you soon.