A body is released from the top of a tower of height h. It takes t sec to r

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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

\frac{t}{2}

s.

Options:

A . At h2 from the ground

B . At h4 from the ground

C . Depends upon mass and volume of the body

D . At 3h4 from the ground

Answer: Option D
:
D
Let the body after time

\frac{t}{2}

fall a distancex from the top. Since it starts from rest, we have

x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}

.............(i)
It reaches the ground in t seconds, we have

h = \frac{1}{2} g t^{2}

..............(ii)
Eliminate t from (i) and (ii), we get x =

\frac{h}{4}

∴

Height of the body from the ground
=

h - \frac{h}{4} = \frac{3 h}{4}

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