A ball is dropped on to a fixed horizontal surface from a height h, the coefficient of restitution is e. The average speed of the ball from the instant it is dropped till it goes to maximum height after first impact with ground

Engineering

Physics

Acceleration

Straight Line Motion

Collision

Question

A ball is dropped on to a fixed horizontal surface from a height h, the coefficient of restitution is e. The average speed of the ball from the instant it is dropped till it goes to maximum height after first impact with ground

$\frac{(1 + e^{2})}{(1 + e)} \sqrt{\frac{gh}{2}}$

$\frac{(1 - e^{2}) \sqrt{2 gh}}{{(1 + e)}^{2}}$

$\frac{(1 - e) \sqrt{2 gh}}{(1 + e)}$

$\frac{(1 + e) \sqrt{2 gh}}{(1 - e)}$

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The average speed is total distance divided by total time. The ball falls from height h, hits ground with speed $\sqrt{2 g h}$ . After first impact, it rebounds with speed $e \sqrt{2 g h}$ and rises to height $e^{2} h$ .

Total distance = h (fall) + $e^{2} h$ (rise) = $h (1 + e^{2})$ .

Total time = time to fall + time to rise = $\sqrt{\frac{2 h}{g}}$ + $e \sqrt{\frac{2 h}{g}}$ = $\sqrt{\frac{2 h}{g}} (1 + e)$ .

Average speed = $\frac{h (1 + e^{2})}{\sqrt{\frac{2 h}{g}} (1 + e)}$ = $\frac{(1 + e^{2})}{(1 + e)} \sqrt{\frac{g h}{2}}$ .

Final Answer: $\frac{(1 + e^{2})}{(1 + e)} \sqrt{\frac{g h}{2}}$