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Deriving max projectile displacement given time

Video transcript
Just want to follow up on the last video, where we threw balls in the air, and saw how long they stayed up in the air. And we used that to figure out how fast we initially threw the ball and how high they went in the air. And in the last video, we did it with specific numbers. In this video, I just want to see if we can derive some interesting formulas so that we can do the computations really fast in our brains, while we're playing this game out on some type of a field, and we don't necessarily have any paper around. So let's say that the ball is in the air for delta t. Delta t is equal to time in the air. Then we know that the time up is going to be half that, which is the same thing as the time down. The time up is going to be equal to delta t-- I'm going to do that in the same color-- is going to be equal to the time in the air divided by 2. So what was our initial velocity? Well, all we have to do is remind ourselves that the change in velocity, which is the same thing as the final velocity minus the initial velocity. So the final velocity-- remember, we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. I won't even write it. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. So let's figure out our total distance, or the distance that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- so it's final velocity is 0. We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. This is the same thing. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation. So this is our max displacement, kind of how high we get displaced. Right when that ball is stationary, or has no net velocity, just for a moment, and starts decelerating downwards. So we can use that. If a ball is in the air for 5 seconds-- we can verify our computation from the last video-- our maximum displacement, 1.225, times 5 squared, which is 25, will give us 30.625. That's what we got in the last video. If the ball's in the air for, I don't know, 2.3 seconds-- so it's 1.225 times 2.3 squared-- then that means it went 6.48 meters in the air. So anyway, I just wanted to give you a simple expression that gives you the maximum displacement from the ground, assuming air resistance is negligible, as a function of the total time in the air. I don't know. I find that pretty fun. And it's a neat game to play.