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?
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