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A Bouncing Ball

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

August 17, 2017

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