Two bodies of different masses $m_{1}$ and $m_{2}$ are dropped from two different heights $h_{1}$ and $h_{2}$ . The ratio of the times taken by the two to reach the ground is

m_{1} : m_{2}

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\sqrt{m_{1}} : \sqrt{m_{2}}

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h_{1} : h_{2}

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\sqrt{h_{1}} : \sqrt{h_{2}}

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Solution

The correct option is D $\sqrt{h_{1}} : \sqrt{h_{2}}$
From second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$

Given, $u = 0, a = g, s_{1} = h_{1}, s_{2} = h_{2}$

$h_{1} = (1 / 2) g (t_{1})^{2}$
$h_{2} = (1 / 2) g (t_{2})^{2}$
So, $h_{1} / h_{2} = (t_{1})^{2} / (t_{2})^{2}$
or $(t_{1}) / (t_{2}) = \frac{\sqrt{h_{1}}}{\sqrt{h_{2}}}$

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