Question

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

Answer 1

Speed of efflux, $v_0=\sqrt{2g(10-h)}$
Component of its velocity parallel to the plane is $v_0\cos {30}^{\circ }$

Let the stream strike the plane after time $t$. Then, along the plane,
$0=v_0\cos {30}^{\circ }-g\sin {30}^{\circ }t$
[Since the stream strikes perpendicular to the plane]
$\Rightarrow t=\frac{v_0\cot {30}^{\circ }}{g}$

Along horizontal direction,
$x=v_{0}t=\frac{v_0^2\cot {30}^{\circ }}{g}----\left(1\right)$
Also, from $\Delta OAB,$
$x=\sqrt{3}(h-y)=\sqrt{3}(h-\frac{1}{2}g{t}^2)$

$\Rightarrow \frac{\sqrt{3}{v}_0^2}{g}=\sqrt{3}(h-\frac{1}{2}\times g\times \frac{v_0^2{\cot }^2{30}^{\circ }}{g^2})$
$\Rightarrow \frac{v_0^2}{g}=h-3\frac{v_0^2}{2g}$
$\Rightarrow h=\frac{5v_0^2}{2g}$
$\Rightarrow h=\frac{5}{2g}\times 2g(10-h)\Rightarrow h=5(10-h)$
$\Rightarrow h=8.33\text{m}$