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A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time $t / 2$ sec

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At h/4 from the ground

At 3h/4 from the ground

At h/2 from the ground

Depends upon mass and volume of the body

answer is D.

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Detailed Solution

$h = \frac{1}{2} {gt}^{2}$

$h^{'} = \frac{1}{2} g \frac{t^{2}}{4} = \frac{h}{4}$ from the top

$= \frac{3 h}{4}$ from the ground

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