(D) Using the second equation of motion,$S = ut + \frac{1}{2}at^2$. Since the bodies are dropped,the initial velocity $u = 0$ and acceleration $a = g$

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(D) Using the second equation of motion,$S = ut + \frac{1}{2}at^2$. Since the bodies are dropped,the initial velocity $u = 0$ and acceleration $a = g$. For the first body,$h_1 = \frac{1}{2}gt_1^2$,which implies $t_1 = \sqrt{\frac{2h_1}{g}}$. For the second body,$h_2 = \frac{1}{2}gt_2^2$,which implies $t_2 = \sqrt{\frac{2h_2}{g}}$. The ratio of the times taken is $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$.

Class 11 Physics Motion in a Straight Line Motion Under Gravity

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Two bodies of masses $m_1$ and $m_2$ are dropped from two different heights $h_1$ and $h_2$ respectively. The ratio of the times taken by the two masses to touch the ground is (neglect air resistance)

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A
$h_1 / h_2$
B
$m_1 h_1 / m_2 h_2$
C
$m_1 h_2 / m_2 h_1$
D
$\sqrt{h_1 / h_2}$

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Two bodies of masses $m_1$ and $m_2$ are dropped from two different heights $h_1$ and $h_2$ respectively. The ratio of the times taken by the two masses to touch the ground is (neglect air resistance)

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