Select Page

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