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Problem: A ball is thrown straight upward so that it reaches a height $h$. It falls down and bounces repeatedly. After each bounce, it returns to a certain fraction $f$ of its previous height. Find the total distance traveled, and also the total time, before it comes to rest. What is its average speed?

Attempt at solution: Let $h$ be the initial height of the ball when it is thrown up. Then a distance of $2h$ is covered before the first bounce. Before the second bounce, a distance of $2hf$ is covered, before the third $2hf^2$, and so on.

Hence we have that $s = 2h + 2hf + 2hf^2 + ...,$ with $s$ the total distance covered. This is a geometric series of the form \begin{align*} \sum_{n=1}^{\infty} a r^{n-1}, \end{align*} with $a = 2h$ and $r = f$ in our case. Since $|f| < 1 $ (it is a fraction), this series converges to \begin{align*} \frac{2h}{1-f} = s. \end{align*}

But now I don't know how to find the total time. From conservation of energy, we have that $mgh = \frac{1}{2} mv^2$, hence $v = \sqrt{2gh}$ at some time. I don't know how to proceed though; any help please?

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  • $\begingroup$ Pretty sure it never really comes to rest... $\endgroup$
    – Hasan Saad
    Mar 22 '15 at 11:24
  • $\begingroup$ In the $n^{th}$ bounce, the ball start with velocity $v = \sqrt{2 g h f^{n-1}}$ pointing up and ends with same velocity but pointing down. The duration of the $n^{th}$ bounce is $$\frac{v - (-v)}{g} = \sqrt{\frac{8h}{g}f^{n-1}}$$ and the total time for all the bounces is $$\sqrt{\frac{8h}{g}}\sum_{n=1}^\infty f^\frac{n-1}{2} = \sqrt{\frac{8h}{g}}\frac{1}{1-\sqrt{f}}$$ $\endgroup$ Mar 22 '15 at 11:42
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Just repeat what you did to find the total distance. You will be able to calculate The time to travel the first 2h distance as $\frac{2\sqrt{2gh}}{g}$ and so on. The sequence of time will also be an geometric progression. $t_{total}=\frac{2\sqrt{2gh}}{g}(1+f^{\frac{2}{2}}+f^{\frac{3}{2}}+f^{\frac{4}{2}}+\ldots)$.

To find the time to travel the first $2h$ just use $v=u+at$. The initial velocity is $\sqrt{2gh}$ as you have mentioned and the final velocity is also that value but oppsite in direction.

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  • $\begingroup$ Thanks for the answer. How did you deduce the $2\sqrt{2gh}/g$ for the first $2h$ distance? $\endgroup$
    – Kamil
    Mar 22 '15 at 12:01
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You can approach the total time in a similar way, by considering the first bounce to height h and the time that one bounce takes. Clearly the time taken will depend on h. For each of the subsequent bounces, use fh or f^2h etc as the height that you now know the dependence of on t

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  • $\begingroup$ Can you show me how please. $\endgroup$
    – Kamil
    Mar 22 '15 at 11:31

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