A ball is dropped from a height \(h\) above a horizontal concrete surface. The coefficient of restitution for the collision involved is \(e\). What is the time after which the ball stops bouncing?

**Discussion:**

The time required for the free fall of the ball is \(\sqrt{\frac{2h}{g}}\). Then the time taken for rise and next fall will be \(2\sqrt{\frac{2h}{g}}e\).

Time taken for one more rise and fall will be \(2\sqrt{\frac{2h}{g}(e^2)}\). Thus, the total time for which the ball will be in motion will be

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

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