First slide

Rectilinear Motion

Question

A body is released from the top of a tower of height h takes time ‘t’ to reach the ground. After time $\frac{' t'}{2}$ , the height from the ground is…..

Easy

A

$\frac{h}{3}$
B

$\frac{h}{2}$
C

$\frac{h}{4}$
D

$\frac{3 h}{4}$

Solution

For a freely falling body $h = \frac{1}{2} g t^{2}$
Let ‘y’ be the distance travelled by it after $\frac{' t'}{2} s$
$∴ y = \frac{1}{2} g {(\frac{t}{2})}^{2}$
$= \frac{1}{4} (\frac{1}{2} g t^{2}) = \frac{h}{4}$
$∴$ The height of the body from the ground is $h^{1} = h - y$
$= h - \frac{h}{4} = \frac{3 h}{4}$ .

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