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# Determine the height above the dashed line XX’ attained by the water stream coming out through the hole situated at point B in the diagram given. Given that $h = 10m$, $L = 2m$, $d = 30^\circ$. (A) 10m(B) 7.1m(C) 5m(D) 3.2m

Last updated date: 13th Jun 2024
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Hint: The velocity of the water stream at exit can be given by Torricelli's equation. Consider the maximum height attained by the water stream.
$\Rightarrow {v^2} = 2gh$ where $v$ is the velocity of stream at exit, $g$ is the acceleration due to gravity and $h$ is the height of the free surface of the liquid from exit,
$\Rightarrow H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}}$ where $H$ is the maximum height of the water stream from exit point and $\alpha$ is the angle the velocity makes with the horizontal.

Torricelli’s equation gives us the velocity of the water stream at exit due to the pressure of the liquid above the exit.
To solve, let us call the height of exit above line XX’ equal to $h'$.
Then, $h' = L\sin \alpha$.
Substituting the known values we get,
$\Rightarrow h' = 2\sin 30^\circ$
$\Rightarrow 2 \times \dfrac{1}{2}$
$\Rightarrow 1m$
The height of exit above the free surface is
$\Rightarrow h - h' = 10 - 1$
$\Rightarrow 9m$
Where $h$ is the height of the free surface of liquid from line XX’.
Then, velocity at the exit is given by
$\Rightarrow {v^2} = 2g(h - h')$
where $v$ is the velocity of stream at exit, $g$ is the acceleration due to gravity.
Therefore, substituting the known values, we get
$\Rightarrow {v^2} = 2\left( {9.8} \right)\left( 9 \right)$
$\Rightarrow 176.4$
Square rooting both sides we get,
$\Rightarrow v = \sqrt {176.4}$
$\Rightarrow 13.3m/s$
Therefore, the velocity at the exit is $13.3m/s$.
The maximum height attained by the water stream will be
$\Rightarrow H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}}$
$\Rightarrow \dfrac{{176.4\left( {{{\sin }^2}30} \right)}}{{2\left( {9.8} \right)}}$
Solving further we get,
$\Rightarrow H = \dfrac{{176.4\left( {{{\left( {\dfrac{1}{2}} \right)}^2}} \right)}}{{2\left( {9.8} \right)}}$
$\Rightarrow \dfrac{{176.4\left( {\dfrac{1}{4}} \right)}}{{2\left( {9.8} \right)}}$
$\Rightarrow 2.25m$
Therefore, the total height from the line XX’ is
$\Rightarrow {H_t} = H + h'$
$\Rightarrow 2.25 + 1$
$\Rightarrow 3.25m$
The closest to the answer is option D.
Hence, our correct option is (D).

Note
A common point of error is to incorrectly leave $H = 2.25m$ as the final answer. Note that the maximum height $H$ of any projectile as calculated from $H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}}$ is only the height of the body from the point where it has the velocity $v$. Hence, for the water stream, this is the exit point which is 1m above line XX’.