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(B) Let the body start from rest at point $A$ with initial velocity $u = 0$.For the motion from $A$ to $B$ (falling through height $h$):Using the equa

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$4$

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(B) Let the body start from rest at point $A$ with initial velocity $u = 0$.For the motion from $A$ to $B$ (falling through height $h$):Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$,$a = g$,and $s = h$:$v^2 = 0 + 2gh$  ---$(i)$For the motion from $A$ to $C$ (falling through total height $x$ to reach velocity $2v$):Using the same equation $v_f^2 = u^2 + 2as$,where $v_f = 2v$,$u = 0$,$a = g$,and $s = x$:$(2v)^2 = 0 + 2gx$$4v^2 = 2gx$  ---(ii)Dividing equation (ii) by equation $(i)$:$\frac{4v^2}{v^2} = \frac{2gx}{2gh}$$4 = \frac{x}{h}$$x = 4h$Therefore,the total distance from the starting point is $4h$.

$A$ body freely falling from rest has a velocity $v$ after it falls through a height $h$. The total distance it has to fall down from the starting point for its velocity to become double is: (in $h$)

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