A water barrel stands on a table of height $h$. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizonatal distance $R$ from the barrel. The depth of water in the barrel is
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$v=\sqrt{2 g d}, R=v t=t \sqrt{2 g d}$
and $h=\frac{1}{2} g t^{2}$
$\therefore h=\frac{1}{2} g \frac{R^{2}}{2 g d}=\frac{R^{2}}{4 d}$
$\Rightarrow d=\frac{R^{2}}{4 h}$
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