A body freely falling from rest has a velocity $v$ after it falls through a distance $h$ . The distance it has to fall down further, for its velocity to become double is:

4 h

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3 h

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h

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16 h

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Solution

The correct option is B $3 h$
$v^{2} = u^{2} + 2 g h$ ....(1)
$v_{1}^{2} = u^{2} + 2 g (h + x)$ ... (2)
$v_{1} = 2 v$
$\Rightarrow 4 v^{2} = 2 g (h + x)$
$8 g h = 2 g (h + x)$
$x = 3 h$

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A body falling from rest has a velocity v after it falls through a height h. The distance it has to fall for its velocity to be 2v is.

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