Coefficient of restitution for bouncing ball

Question

If I throw a bouncy ball at the ground very hard, it may rebound to a height greater than that from which it was thrown.

The coefficient of restitution is given by $$e=\dfrac{\text{speed of separation}}{\text{speed of approach}},$$ and I recall learning at school (several years ago) that $0\leqslant e\leqslant 1$. But this implies the speed of separation is always at most the speed of approach. Assuming the ball is thrown vertically downwards and rebounds vertically upwards, $v^2=u^2+2as$ gives us the rebound height, $s$: $$s=\dfrac{u^2}{2g}\,,$$ where $u$ is the speed of separation.

Now let $h$ be the height from which the ball was dropped. Then $v^2=2gh$, where $v$ is the speed of approach. Since $u\leqslant v$, we have $$s=\dfrac{u^2}{2g}\leqslant \dfrac{v^2}{2g}=\dfrac{2gh}{2g}=h.$$

But for some bouncy balls, this is not true, which appears to imply $e>1$. According to the Wikipedia page for coefficient of restitution, $e>1$ represents collisions in which energy is released, such as things which explode at the point of impact.

Is $e>1$ in the case of some bouncing balls? If so, is this because of some change in energy?

And when is it reasonable to assume $0\leqslant e\leqslant 1$? I think lots of elementary mechanics textbooks do this.

Answer 1

When you make the statement:

But this implies the speed of separation is always at most the speed of approach

you are correct.

However, if you look carefully in what you have said, you are bringing two scenarios into play that cannot both be true at the same time. At the very beginning of the post you say:

If I throw a bouncy ball at the ground very hard . . .

But then later you say:

Now let h be the height from which the ball was dropped. Then v2 = 2gh, where v is the speed of approach

A ball cannot be dropped from rest as well as thrown. The equation v2=2gh only applies to balls that were dropped from rest. It is out of the question once you declare that the ball was thrown down very hard.