A tank is filled up to a height $2H$ with a liquid and is placedon a platform of height $H$ from the ground. The distance $x$ from the ground where a small hole is punched to get the maximum range $R$ is:
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$v=\sqrt{2 g(3 H-x)}, R=v \cdot t$
and $x=\frac{1}{2} g t^{2}$
$\therefore R=\sqrt{2 g(3 H-x)} \sqrt{\frac{2 x}{g}}$
$=2 \sqrt{\left(3 H x-x^{2}\right)}$
For $R=\max , \frac{d R}{d x}=0$
$=\frac{d}{d x} \sqrt{3 H x-x^{2}}$
$=\frac{1}{2 \sqrt{3 H x-x^{2}}}(3 H-2 x)$
$\Rightarrow x=\frac{3}{2} H=1.5 H$
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