Question about dropping a ball and coefficient of restitution.

Question

If I drop a ball from a height $h$ and the ball rebounds from the floor it will bounce back up to a height of $e^2h$ where $e$ is the coefficient of restitution between the floor and the ball. Why is this the case?

Answer 1

Assuming air friction is negligible, the speed of the ball before impact is $\sqrt{2gh}$. If the ball bounces back up to a height $d$, then the speed of the ball after impact is $\sqrt{2gd}$. Therefore, the coefficient of restitution is

$$e = \frac{\sqrt{2gd}}{\sqrt{2gh}} = \sqrt{\frac{d}{h}}.$$

Therefore $d = e^2 h$.

Answer 2

Well, you could do it another way. Explicitly, it could be stated $f(e)h$, where $f(e)$ is a function of e. Also, $e^2$ could be simplified to just a constant $c$. However, this constant may have other properties useful in say a spring system. In one sentance, the number e may be used in other formulas, so instead of having additional constants for springs and other systems you just make one constant and convert it, and put it in a function for the system your looking at.