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(A) For a body dropped from a height $h$,the time taken to reach the ground is given by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the i

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$1:\sqrt{2}$

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(A) For a body dropped from a height $h$,the time taken to reach the ground is given by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the initial velocity $u = 0$,we have $h = \frac{1}{2}gt^2$,which gives $t = \sqrt{\frac{2h}{g}}$.For the first ball dropped from height $h_1 = h$,the time taken is $t_1 = \sqrt{\frac{2h}{g}}$.For the second ball dropped from height $h_2 = 2h$,the time taken is $t_2 = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}}$.The ratio of the times is $\frac{t_1}{t_2} = \frac{\sqrt{2h/g}}{\sqrt{4h/g}} = \sqrt{\frac{2h}{4h}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,the ratio is $1:\sqrt{2}$.

Two balls are dropped from heights $h$ and $2h$ respectively from the Earth's surface. The ratio of the time taken by these balls to reach the Earth is

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