Two bodies of different masses $m_{a}$ and $m_{b}$ are dropped from two different height $a$ and $b$ . The ratio of the time taken by the two to cover these distances are

a : b

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b : a

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\sqrt{a} : \sqrt{b}

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a^{2} : b^{2}

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Solution

The correct option is C $\sqrt{a} : \sqrt{b}$
Since initial velocity is zero, we can write
$h = \frac{1}{2} g t^{2}$ (Taking downward direction as positive) $t = \sqrt{\frac{2 h}{g}}$
$t_{a} = \sqrt{\frac{2 a}{g}}$
$t_{b} = \sqrt{\frac{2 b}{g}}$
$\frac{t_{a}}{t_{b}} = \sqrt{\frac{a}{b}}$

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