Question

A ball is dropped from a height $h$ on a floor, where the coefficient of restitution is $e$. Find the time interval between the instant of drop and second collision with ground.

• A. $\sqrt{\frac{2h}{g}}(1+e)$
• B. $\sqrt{\frac{2h}{g}}(1+\frac{e}{2})$
• C. $\sqrt{\frac{2h}{g}}(1+2e)$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{2h}{g}}(1+2e)$

Time taken by the ball to reach the ground after it is dropped,

$t_1=\sqrt{\frac{2h}{g}}$

Upward velocity immediately after the first collision,

$v=e\sqrt{2gh}$

The time interval between first and second collision is,

$t_2=2e\sqrt{\frac{2h}{g}}$

$\therefore$ Total time is,

$t=t_1+t_2=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}=\sqrt{\frac{2h}{g}}(1+2e)$

Hence, option $\left(C\right)$ is the correct answer.

Why this question ? Concept - After each rebound, the ball doesn't rise to the same height due to the inelastic collision. Hence, it ultimately comes to rest.