Question

A container is filled with liquid upto height $H$ as shown in figure. There are $2$ holes in the container such that water coming out from them have same range. If the range of water coming out of the holes is half of maximum range, then find the distance $x$ between these holes.

Answer 1

Range $=2\sqrt{(H-h)h}$

So, $R_{max}=H$

According to question,

$R=\frac{H}{2}=2\sqrt{(H-h)h}$

$\Rightarrow \frac{H}{4}=\sqrt{(H-h)h}$

$\Rightarrow \frac{H^2}{16}=(H-h)h$

$\Rightarrow \frac{H^2}{16}-Hh+h^2=0$

$\Rightarrow h=\frac{1\pm \sqrt{1-4\times 1\times \frac{1}{16}}}{2}H=\frac{1\pm \frac{\sqrt{3}}{2}}{2}H=(\frac{1}{2}-\frac{\sqrt{3}}{4})H=0.067H$

and $x+2h=H$

$\Rightarrow x=H-2h=0.866H$