Question

A rectangular tank of height $10m$ filled with water, is placed near the bottom of a plane inclined at an angle ${30}^{\circ }$ with horizontal. At height h from bottom a small hole is made (as shown in figure) such that the stream coming out from hole, strikes the inclined plane normally. Calculate $h$

Answer 1

$v=\sqrt{2g(10-h)}$
Component of its velocity parallel to the plane is $v\cos {30}^{\circ }$.
Let the stream strikes the plane after time t. Then
$0=v\cos {30}^{\circ }-g\sin {30}^{\circ }t$
$\therefore$ $t=\frac{v\cos {30}^{\circ }}{g}$
Further $x=vt=\frac{v^2\cos {30}^{\circ }}{g}$
$=\sqrt{3}y$
or $\frac{v^2\cos {30}^{\circ }}{g}=\sqrt{3}(h-\frac{1}{2}g{t}^2)$
$\therefore$ $\frac{\sqrt{3}{v}_2}{g}=\sqrt{3}(h-\frac{g}{2}\frac{v^2{\cos }^2{30}^{\circ }}{g^2})$
or $\frac{v^2}{g}=h-\frac{3}{2}\frac{v^2}{g}$
$\therefore$ $\frac{5}{2}\frac{v^2}{g}=h$
or $5(10-h)=h$
$\therefore$ $h=8.33m$