A rubber ball is dropped from a height of 5 | vedclass

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(B) Let the ball fall from a height $h_1$ and reach a height $h_2$ after bouncing. The coefficient of restitution $e$ is given by $e = \sqrt{\frac{h_2

Vedclass · Accepted Answer

$2/5$

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(B) Let the ball fall from a height $h_1$ and reach a height $h_2$ after bouncing. The coefficient of restitution $e$ is given by $e = \sqrt{\frac{h_2}{h_1}}$.Similarly,if the velocities of the ball just before and after impact are $v_1$ and $v_2$ respectively,then $e = \frac{v_2}{v_1}$.Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{1.8}{5}} = \sqrt{\frac{18}{50}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The fraction of velocity lost is given by $\frac{v_1 - v_2}{v_1} = 1 - \frac{v_2}{v_1}$.Substituting the values,we get $1 - \frac{3}{5} = \frac{2}{5}$.

$A$ rubber ball is dropped from a height of $5 \ m$ on a planet where the acceleration due to gravity is unknown. After bouncing,the ball reaches a height of $1.8 \ m$. What fraction of its velocity does the ball lose upon impact?

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