A ball dropped from one metre above the top of a window, crosses the window in $t_{1} s$ . If the same ball is dropped from $2 m$ above the top of the same window, time taken by it to cross the window is $t_{2} s$ . Then

t_{2} = t_{1}

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t_{2} = 2 t_{1}

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t_{2} > t_{1}

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t_{2} < t_{1}

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Solution

The correct option is D $t_{2} < t_{1}$
Solution 1
The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:
$v^{2} = u^{2} + 2 a S$ , where $u = 0$

The ball with higher speed will take less time to cross the window.

$∵$ $H_{2} > H_{1}$
$⟹$ $v_{2} > v_{1}$
$∴$ $t_{2} < t_{1}$
Hence Option D is correct

Solution 2
Statement D is correct
Situation 1:
We want time to cross x.
$v^{2} = u^{2} + 2 a s$
$u = 0, a = - g, s = - 1$
$v^{2} = 2 g \Rightarrow v_{1} = \sqrt{2 g}$
$(1 + x) = \frac{1}{2} g t^{2} \Rightarrow T_{2} = \sqrt{\frac{2 (1 + x)}{g}}$
$1 = \frac{1}{2} g t^{2} = T_{1} = \sqrt{\frac{2}{g}}$
So time taken to cross $x = t_{1} = T_{2} - T_{1} = \sqrt{\frac{2}{g}} (\sqrt{1 + x} - 1)$
Situation 2:
$(2 + x) = \frac{1}{2} g t^{2} \Rightarrow T_{2} = \sqrt{\frac{2 (2 + x)}{g}}$
$2 = \frac{1}{2} g t^{2} \Rightarrow T_{1} = \sqrt{\frac{2 \times 2}{g}}$
So time taken to cross $x = t_{2} = T_{2} - T_{1} = \sqrt{\frac{2}{g}} (\sqrt{2 + x} - \sqrt{2})$
Clearly $t_{2} < t_{1}$

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