Q 11-05-011NEETJEE MainEAMCET 2008 (Engineering)Medium
A ball is dropped from a height $h$ on to a floor of coefficient of restitution $e$. The total distance covered by the ball just before the second hit is
Answer: (B) $h(1 + 2e^2)$
The ball falls $h$ and hits the floor with speed $v = \sqrt{2gh}$. It rebounds with speed $ev$, so it rises to
$$h_1 = \frac{(ev)^2}{2g} = e^2h$$
Before the second hit it goes up $e^2h$ and comes back down $e^2h$:
$$\text{Total distance} = h + 2e^2h = h(1 + 2e^2)$$
Solution by Sreeraj P, M.Sc Physics