Question

A rectangular tank of height $10\,m$ filled with water, 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 figure) such that the stream coming out from holestrikes the inclined plane normally. Calculate $h$

Answer 1

Here, we have to calculate the value of $h$, that is, the height at which the hole is made. For this, we will first calculate the time $t$ at which the water stream strikes the inclined plane. The formula used for calculating the value time t is given below.

Formula used:
Here we will use the formula which is given below
$v = u + at$
Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time taken.

Complete step by step answer:
According to the given question, the value of the velocity will be
$v = \sqrt {2g\left( {10 - h} \right)}$
Here, $v$ is the velocity, $g$ is the acceleration due to gravity and $h$ is the height of the container.
Now, the component of the velocity parallel to the plane will be $v\,\cos 30^\circ$ and the component of the velocity perpendicular to the plane will be $v\,\sin 30^\circ$.When the stream strikes the plane after time $t$. Therefore, the formula of the velocity is given by
$v = u + at$
$\Rightarrow \,0 = v\cos 30^\circ - gt$
$\Rightarrow \,t = \dfrac{{v\cos 30^\circ }}{{g\sin 30^\circ }}$
$\Rightarrow \,t = \dfrac{{v\cot 30^\circ }}{g}$
Now, the wave along the x-axis is represented by the following formula
$x = vt$
$\Rightarrow \,x = \dfrac{{{v^2}\cot 30^\circ }}{g} = \sqrt 3 y$
$\Rightarrow \,\dfrac{{{v^2}\cot 30^\circ }}{g} = \sqrt 3 \left( {h - \dfrac{1}{2}g{t^2}} \right)$
$\Rightarrow \,\dfrac{{\sqrt 3 {v^2}}}{g} = \sqrt 3 \left( {h - \dfrac{1}{2}g{t^2}} \right)$
$\Rightarrow \,\dfrac{{\sqrt 3 {v^2}}}{g} = \sqrt 3 \left( {h - \dfrac{1}{2}g\dfrac{{{v^2}{{\cot }^2}30^\circ }}{{{g^2}}}} \right)$
$\Rightarrow \,\dfrac{{{v^2}}}{g} = \left( {h - \dfrac{3}{2}\dfrac{{{v^2}}}{g}} \right)$
$\Rightarrow \,h = \dfrac{{{v^2}}}{g} + \dfrac{{3{v^2}}}{{2g}}$
$\Rightarrow \,h = \dfrac{{5{v^2}}}{{2g}}$
$\Rightarrow \,h = \dfrac{5}{{2g}} \times 2g\left( {10 - h} \right)$
$\Rightarrow \,h = 5\left( {10 - h} \right)$
$\Rightarrow \,h + 5h = 50$
$\Rightarrow \,6h = 50$
$\therefore \,h = 8.33\,m$

Therefore, the height $h$ will be $8.33\,m$.

Note: The value of the velocity $v = \sqrt {2g\left( {10 - h} \right)}$ is of the water stream that is flowing out of the tank. Also, remember that here final velocity is not given, that is why, we have taken $v = 0$. Also, $v\sin 30^\circ$ is the component of velocity in x-direction and $\sin 30^\circ$ is the component of velocity along y-direction.