A ball is dropped from rest at height 4h. After it has fallen a distance d, a second ball is dropped from rest at height h. What should d be (in terms of h) so that the balls hit the ground at the same time?

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$d = \frac{h}{3}$

$d = h$

$d = \frac{3 h}{2}$

$d = 3 h$

answer is C.

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Detailed Solution

For the 1st ball, $4 h = \frac{{gt}_{1}^{2}}{2} \Rightarrow t_{1} = \sqrt[2]{\frac{2 h}{g}}$ & For second ball $t_{2} = \sqrt[2]{\frac{2 h}{g}}$ also
$\sqrt{\frac{2 h}{g}} + \sqrt{\frac{2 d}{g}} = 2 \sqrt{\frac{2 h}{g}} \Rightarrow \sqrt[2]{h} = \sqrt{d} + \sqrt{h} \Rightarrow d = h$

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