Try this problem, useful for Physics Olympiad, based on a bouncing ball.

The Problem:

A ball is dropped from a height (h) above a horizontal concrete surface. The coefficient of restitution for the collision involved is (e). What is the time after which the ball stops bouncing?

Discussion:

The time required for the free fall of the ball is (\sqrt{\frac{2h}{g}}). Then the time taken for rise and next fall will be (2\sqrt{\frac{2h}{g}}e).
Time taken for one more rise and fall will be (2\sqrt{\frac{2h}{g}(e^2)}). Thus, the total time for which the ball will be in motion will be
$$ \sqrt{\frac{2h}{g}}+2\sqrt{\frac{2h}{g}e(1+e+e^2+….)}$$ $$=\sqrt{\frac{2h}{g}(1+\frac{2e}{1+e})}$$ $$=\sqrt{\frac{2h}{g}(\frac{1-e}{1+e})}$$