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Q.

A highly elastic ball is released from rest at a distance above the ground and bounces up and down. With each bounce a fraction f = 1/10 of its kinetic energy is lost. The time for which the ball will continue to bounce is nearly( take g = 10m/s², h = 5m)

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a

1.86 sec

b

1 sec

c

38 sec

d

84.3 sec

answer is B.

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Detailed Solution

$\begin{array}{l} t = \sqrt{\frac{2 h}{g}} (\frac{1 + e}{1 - e}); K e = \frac{1}{2} m v^{2} \\ K f = \frac{9}{10} (\frac{1}{2} m v^{2}); v_{2} = e v_{1} \\ \sqrt{\frac{9}{10}} v = e . v \Rightarrow e = \sqrt{\frac{9}{10}} \\ t = \sqrt{\frac{2 \times 5}{10}} (\frac{1 + \sqrt{\frac{9}{10}}}{1 - \sqrt{\frac{9}{10}}}) = {(\sqrt{10} + \sqrt{9})}^{2} = 38 \sec \end{array}$

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