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(C) For a body falling from rest,the velocity $v$ after falling a distance $h$ is given by the equation of motion $v^2 = u^2 + 2as$. Here $u = 0$,$a =

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$3$

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(C) For a body falling from rest,the velocity $v$ after falling a distance $h$ is given by the equation of motion $v^2 = u^2 + 2as$. Here $u = 0$,$a = g$,and $s = h$.So,$V^2 = 0 + 2gh$ ... $(i)$Let the total distance covered to acquire a velocity $2V$ be $x$. Using the same equation:$(2V)^2 = 0 + 2gx$$4V^2 = 2gx$ ... (ii)Dividing equation (ii) by equation $(i)$:$4V^2 / V^2 = (2gx) / (2gh)$$4 = x / h$$x = 4h$The total distance to reach velocity $2V$ is $4h$. The distance it must fall *further* from point $B$ is $x - h = 4h - h = 3h$.

$A$ body is dropped from rest. It acquires a velocity $V$ after falling through a distance $h$. What distance must it fall further to acquire a velocity of $2V$ (in $h$)?

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