A ball is released from height $h$ and another ball from height $2 h$ . The ratio of time taken by the two balls to reach the ground is

1 : \sqrt{2}

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\sqrt{2} : 1

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2 : 1

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1 : 2

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Solution

The correct option is A $1 : \sqrt{2}$
By second equation of motion
$S = u t + \frac{1}{2} a t^{2}$
For first ball
$h = \frac{1}{2} g t_{1}^{2}$
$\Rightarrow t_{1} = \sqrt{\frac{2 h}{g}}$
For second ball
$2 h = \frac{1}{2} g t_{2}^{2}$
$\Rightarrow t_{2} = \sqrt{\frac{4 h}{g}}$
$∴ t_{1} : t_{2} = 1 : \sqrt{2}$

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