A ball is released from a height h . If t 1 and t 2 be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t 1 and t 2 .

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TestHub · Accepted Answer

As the body is dropped from a certain height $h$ , its initial velocity will be zero.
For first half, $h_{1} = \frac{h}{2}$ .
Using second equation of motion, $s = u t + \frac{1}{2} a t^{2}$
We have, $\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$ .
Time taken in travel first half of the height, $t_{1} = \sqrt{\frac{h}{g}} . . . (1)$ .
For total height $h$ , using again second equation of motion,
$h = \frac{1}{2} g {(t_{1} + t_{2})}^{2} \Rightarrow (t_{1} + t_{2}) = \sqrt{\frac{2 h}{g}} . . . (2)$
From equation $(1)$ and $(2)$ ,
$\Rightarrow \frac{1}{\sqrt{2}} = \frac{t_{1}}{t_{1} + t_{2}}$
$\Rightarrow 1 + \frac{t_{2}}{t_{1}} = \sqrt{2}$
$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{\sqrt{2} - 1}$
Thus, the relation is $t_{2} = (\sqrt{2} - 1) t_{1}$ .

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