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(A) When a ball is released from height $h$, the time taken to reach the ground is given by the equation of motion $h = \frac{1}{2}gt^2$, which gives

Vedclass · Accepted Answer

$\frac{1}{2}\sqrt{\frac{g}{2h}}$

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(A) When a ball is released from height $h$, the time taken to reach the ground is given by the equation of motion $h = \frac{1}{2}gt^2$, which gives $t = \sqrt{\frac{2h}{g}}$.Since the collision is perfectly elastic, the ball rebounds with the same speed and reaches the same height $h$.The time taken for the return journey is also $t = \sqrt{\frac{2h}{g}}$.The total time period $T$ of one complete oscillation (up and down) is $T = t + t = 2\sqrt{\frac{2h}{g}}$.The frequency $f$ is the reciprocal of the time period: $f = \frac{1}{T} = \frac{1}{2\sqrt{\frac{2h}{g}}} = \frac{1}{2}\sqrt{\frac{g}{2h}}$.

$A$ ball is released from height $h$ which makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is ($g$ = acceleration due to gravity).

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