Question

A ball is released from a height h above the ground. The ball makes multiple collisions with the floor. The total tine taken by the ball so that it comes to stop pernanently. (Coefficient of restitution is e)

• A. $\frac{1+e^2}{1-e^2}\sqrt{\frac{2h}{g}}$
• B. $n\sqrt{\frac{2h}{g}}$
• C. $\frac{1-e}{1+e}\sqrt{\frac{2h}{g}}$
• D. $\frac{1+e}{1-e}\sqrt{\frac{2h}{g}}$

Answer 1

The correct option is D $\frac{1+e}{1-e}\sqrt{\frac{2h}{g}}$

Time taken by ball to get to ground from height $h$, $t=\sqrt{\frac{2h}{g}}$

Velocity of ball when it's about to hit floor, using $v^2-u^2=2gh$

$v=\sqrt{2gh}$

Velocity of ball just after hitting the floor, $v^{\prime }=ev=e\sqrt{2gh}$

Now, time taken by ball to get to maximum height and come back will be equal, so time taken to get to maximum height, $t^{\prime }$,

Using, $v=u-gt$

$t^{\prime }=\frac{e\sqrt{2gh}}{g}=e\sqrt{\frac{2h}{g}}$

Similarly for next round, $t^{\prime \prime }=e^2\sqrt{\frac{2h}{g}}$

Total time, $T=\sqrt{\frac{2h}{g}}+$ $2e\sqrt{\frac{2h}{g}}+$ $2e^2\sqrt{\frac{2h}{g}}+...$

Solving the infinite GP,

$T=\sqrt{\frac{2h}{g}}+\frac{2e\sqrt{2h/g}}{1-e}$

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