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

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

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

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m_{1} h_{2} : m_{2} h_{1}

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m_{1} h_{1} : m_{2} h_{2}

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Solution

The correct option is D $\sqrt{h_{1}} : \sqrt{h_{2}}$
Initial velocity of the stone
$u = o$
Acceleration
$a = g$
Let the height of the tower be $h$ .
Using second equation of motion-
$h = u t + \frac{1}{2} g t^{2}$

$∴$ $h = 0 + \frac{1}{2} g t^{2}$

$\Rightarrow t = \sqrt{\frac{2 h}{g}}$
We get,
$t \propto \sqrt{h}$

$\Rightarrow t_{1} : t_{2} = \sqrt{h_{1}} : \sqrt{h_{2}}$

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