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A ball is thrown straight upward. At 4.00 $\mathrm{m}$ above its launch point, the ball's speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

$y_{\max }=5.33 \mathrm{m}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

Lectures

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Okay, So in this problem we have a ball that is thrown vertically, upwards and when it is four meters above the ground. So me start recording this information when we are four meters above the ground, its velocity is half of what it started with. So this initial velocity is going to be twice as much as that final velocity. Okay. And we also know that we're on the planet Earth. So my acceleration is negative. 9.8 meters per second squared. Now, in this problem here, our goal is to figure out how high the ball will eventually reach if it is at half speed of four meters in the air. Um, which means we're actually gonna need to do this in two steps. Because in order to answer that question of how high will it get? We first need to know how fast was it actually going when it left the ground. So that means looking at my numbers here. Um, I have one unknown, and that's that velocity. Ah, because I have three numbers I'm looking with or working with, and the one variable we don't know and aren't looking for is time. So I'm gonna be using my Kinnah Matic equation over here on the right. That does not have anything to do with time. And that means my equation is V squared is equal to be not squared. Plus two a axe. When I plug, my number's in here. That final velocity of 1/2 of the becomes 1/4 because that half gets squared as well. V squared equals V squared plus a two times a negative 9.8 times of four gives us a negative 78.4. Now, when I tell you this help a bit since I have to be squared on both sides, I can combine my leg terms. And then since both sides are negative, we can straighten that up a little bit and we end up with 3/4 of that velocity, um, squared is equal to 78.4 and then I will multiply both sides by 4/3 to cancel that fraction. We have these squared equals 104.5 and take the square root to show that our final velocity is 10.2 meters per second. Now that again is the first part of the problem because this tells us how fast the ball is initially going. Our end goal here is to figure out how fast will it be going or not? How fast, How high up it will actually get when it comes to a complete stop. Um, to me dribble more of my tables in there. There we go. So we need to do one more of these problems and this one's gonna be really similar to what we just did, because the initial velocity is our answer from before. The initial velocity is 10.2 meters per second. The acceleration is still negative. 9.8 meters per second squared. Only peace that's different here is we don't know what exes. That's actually what we're trying to find. But we do know this ball, uh, rises to its maximum height, so its final velocity is going to be zero meters per second. We're looking for how high up it is when we reach zero meters per second. Again, we don't know time, nor do we care about time. So we're going to use that same equation. V squared equals V not squared, plus two a x. When I plug in my numbers. Here we have zero squared, which is zero is equal to the not squared. So 10.2 squared plus two a x. So multiply those together to get negative 19.6 x. Then I'm gonna add that over and take the school and square that number. So we get 19.6 x equals 104.4 Divide both sides by 19.6 and I get my final answer. The maximum height this thing will reaches 5.31 meters off the ground.

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