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Bouncing ball for time Suppose a rubber ball, when dropped from a given height, returns to a fraction $p$ of that height. In the absence of air resistance, a ball dropped from a height $h$ requires $\sqrt{2 h / g}$ seconds to fall to the ground, where $g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration due to gravity. The time taken to bounce up to a given height equals the time to fall from that height to the ground. How long does it take for a ball dropped from 10 m to come to rest?

$$\sqrt{\frac{20}{g}}\left(\frac{1+\sqrt{p}}{1-\sqrt{p}}\right)$$

Calculus 2 / BC

Chapter 9

Sequences and Infinite Series

Section 3

Infinite Series

Sequences

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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Hi. So in this question were pretty much looking at how long it takes for a ball to come. Teoh rest after being bounced from a height of 10 meters. So our initial height is gonna be 10 meters p is the fraction of that initial height. Um, the time, both ways. So when the ball is bouncing up, Andi falling down are both equal to the square root of two times H over G. And our acceleration is gravity. Pretty much so, Um, those are the values that we have right now. So first thing we're gonna dio as we can, um, find out, um, ignoring for right now the initial drop. Um, we can easily figure out what the total time for the end bounce was. So the first thing we have to dio is figure out what the bounce would be, what that would look like. So we know that the initial is 10. So our initial list 10 and any bounce after that would be times P because peas a fraction of that, so we would have tend to the P. And then it would continue. Of course, if we tend to the P um squared on, then it would be 10 p cube. And that would just keep on continuing until we get 10 to the P to the end. So this would be for our end, um, bounce from 10 meters. That would be the height. So we would know Then, Um what are height is for end. So I'm going to just say h on is equal to 10. PM on. We know that the amount of time it takes is for the total going, um, down and then up or up and down in this case was gonna be up and then down to that height, um, it would be two times the number we have right here. So what we can say is our time for end. I'm going to say t n is going to be eat Teoh too high. We're root of two. And this case, we're gonna put our h and here divided by G. So this is really just equal Teoh two times the square root of two times 10 times p to the end over G. So now what we can do is we can calculate the total time because we have an end formula which will make it much easier to figure out how long it takes before the ball comes to arrest from 10 meters. So we know what our initial amount of time would be. Um, because it's just given by this formula. So our initial would Onley be the amount of time it takes for the ball to drop the ball will never again reached that height to stock going up to that height. So our initial would just be the square root of two times H zero over G, which is equal to the square root of two times 10. Cause that is our initial over G, which is just gonna be equal to the square root of 20 over J. And now we can use both of these pieces of information in order to form an equation. So we have our initial and we're going to say that it's 20 over g. It's gonna be plus a Siri's. So we're gonna have Infiniti. We're gonna have a new equal, uh, 21 We're going to say, um, that that is going to be of to times to times 10 to the p on over G. So just using our formula up here that we have because we're trying to find total monotypes. We have our initial time, and then this is the amount for up to and our end bats. Now, from here, we can simplify our equation of it. Weaken Be right. This so we can write this as 20 over G. Um, what? And now what we can do is we can, um, again we'll have our signal here for now. But we're going to just take our two out, and we Can we write this as being 20 over g? Um, on That is gonna be times p on in there. So from here, what we can do is we'll have our 20 over G. We're gonna add that again to our Siri's. But we're gonna do is we're gonna separate these two. So we're gonna have R P. And and now from here we're gonna do is recording too. Take out, um, are 20 g from inside Siri's. We're just going to write it on the outside. It does not have any, um, variables that are needed in this case. So doesn't have to be in the Siri's. It's just a constant, um, on now from here we're going to Dio is we're going to put this equation into formula. That I'm sure you've seen a lot in this chapter, which is a sorry. A over one minus are. So as we wrote up here, um, we know that we could take out our 10 um, and knees equations. It's not necessary. But r r P is really are Siri's here. So we have p and then we have plus piece squared. Plus, he cubed, continuing on to pee to the at, um so we can see that r p is really where are Siri's is gonna come from, and right now we have it as the square root of Peter the end. So we do need to write out our new Siri's. So we're gonna have, uh, just pee, and then it's going to be Plus, he squared plus cubes. So what we can see is that are a value is going to be the 1st 1 severe a value. And now we need to calculate our our value, which is just going to be our 2nd 1 divided by first. So it's just going to be equal to the square root of P. So now we can rewrite. Oh, this'll down here. Actually, Onley going to be rewriting this part and then we're going to put the other part into our handy formula over here. So we're going to rewrite this as 20 over G plus two times 20 over G, and then we have our A value is a square root of P, and that is over one minus the square root of P. And from here, we're just going to start simplifying our equations. Um, so right now we have, um, to parts that have the square root of 20 over g so we can just take out our 20 over she on. That's just be a one here and then plus to times are Formula Square with P over one, minus p on from here. Um, we can simplify even more because we can say that this is gonna be equal 21 minus the square root of p over one minus the square root of P class to times the square root of P over one minus the square root of P on. This is then equal to square root of 20 over g times, one plus the square root p over one minus the square root of P. Andi, that is gonna be our answer. This is Thea. Amount of seconds. So it takes this money seconds before the ball will come to arrest.

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