Question

A ball is dropped from height $H$ on to a horizontal surface. If the coefficient of restitution is $e$ then the total time after which it comes to rest is

• A. $\sqrt{\frac{2 H}{g}}\left(\frac{1 - \, e}{1 + \, e}\right)$
• B. $\sqrt{\frac{2 H}{g}}\left(\frac{1 + e}{1 - e}\right)$
• C. $\sqrt{\frac{2 H}{g}}\left(\frac{1 + e^{2}}{1 - \, e^{2}}\right)$
• D. $\sqrt{\frac{2 H}{g}}\left(\frac{1 - \, e^{2}}{1 + e^{2}}\right)$

Answer 1

Answer: (B)

$\sqrt{\frac{2 H}{g}}\left(\frac{1 + e}{1 - e}\right)$

Answer 2

The time it takes for the ball to fall from a height is given by $t=\sqrt{\frac{2 H}{g}}$ The height up to which the ball bounces after the $n^{th}$ collision from the floor is $H_{n}=e^{2 n}H$ $t=\sqrt{\frac{2 H}{g}}+2\sqrt{\frac{2 H_{1}}{g}}+2\sqrt{\frac{2 H_{2}}{g}}+\ldots \infty$ $t=\sqrt{\frac{2 H}{g}}\left(1 + 2 e + 2 e^{2} + \, \ldots \ldots . . \in fty\right)$ $t=$ $\sqrt{\frac{2 H}{g}}\left(\frac{1 + e}{1 - e}\right)$