A rubber ball dropped from a height of $1 m$ bounces (after hitting the floor) several times, losing $10 %$ of its $K . E .$ in each after bounce. Find the speed of the ball after the $n - t h$ bounce:

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Solution

By conservation of mass
$m g h = \frac{1}{2} m v^{2}$
$v = \sqrt{2 g h} = \sqrt{20} m / s$
$\frac{1}{2} m v_{1}^{2} = \frac{1}{4} m v^{2}$
$v_{1} = \frac{3 v}{\sqrt{10}}$
Similarly
$v_{2} = v (\frac{3}{\sqrt{10}})^{2}$
$v_{n} = v (\frac{3}{\sqrt{10}})^{n}$
$v_{n} = \sqrt{20} (\frac{3}{\sqrt{10}})^{n}$

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A rubber ball dropped from a height of 1 m bounces (after hitting the floor) several times, losing 10% of its K.E. in each after bounce. Find the speed of the ball after the n−th bounce:

By conservation of massmgh=12mv2v=√2gh=√20m/s12mv21=14mv2v1=3v√10Similarlyv2=v(3√10)2vn=v(3√10)nvn=√20(3√10)n

A rubber ball dropped from a height of $1 m$ bounces (after hitting the floor) several times, losing $10 %$ of its $K . E .$ in each after bounce. Find the speed of the ball after the $n - t h$ bounce: