A body dropped from a certain height attains the same velocity as another falling with an initial velocity u from a height h below the first body. If g is the acceleration due to gravity, then

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$u = \sqrt{g h}$

$u = \sqrt{2 g h}$

$u = \sqrt{\frac{g h}{2}}$

$u = 2 \sqrt{g h}$

answer is C.

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Detailed Solution

$υ_{A}^{2} = 2 g (h + x)$ and $υ_{B}^{2} = u^{2} + 2 g x$
Since $υ_{A} = υ_{B}$
$\Rightarrow u^{2} + 2 g x = 2 g h + 2 g x$
$\Rightarrow u^{2} = 2 g h$
$\Rightarrow u = \sqrt{2 g h}$

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