Question

A hose lying on the ground shoots a stream of water upward at an angle of ${{60}^{\circ }}$ to the horizontal with the velocity of $16m{{s}^{-1}}$. The height at which the water strikes the wall 8m away is :
A. 8.9m
B. 10.9m
C. 12.9m
D. 6.9m

Answer 1

First calculate the initial horizontal and vertical speeds of the projectile. Find the time taken by the water stream to reach the wall which is 8m away. Then calculate the vertical position of the projectile at this time.

Formula used:
${{u}_{x}}=u\cos \theta$
${{u}_{y}}=u\sin \theta$,
where ${{u}_{x}}$, ${{u}_{y}}$ are the initial horizontal and vertical speed of the projectile, u is initial speed and $\theta$ is angle of projection.
$t=\dfrac{d}{v}$,
t is time, d is distance covered and v is speed.
$s=ut+\dfrac{1}{2}a{{t}^{2}}$,
where s, u, t and a are the displacement, initial velocity, time and acceleration of a body moving in a straight line.

Complete step by step answer:
The given case is of a projectile motion. The projectile is projected with an initial velocity of $16m{{s}^{-1}}$ making an angle of ${{60}^{\circ }}$ with the horizontal.
This means $u=16m{{s}^{-1}}$ and $\theta ={{60}^{\circ }}$.
Then this implies that,
${{u}_{x}}=u\cos \theta \\\ \Rightarrow{{u}_{x}} =16\cos {{60}^{\circ }}\\\ \Rightarrow{{u}_{x}} =16\times \dfrac{1}{2}\\\ \Rightarrow{{u}_{x}} =8m{{s}^{-1}}$.
And
${{u}_{y}}=u\sin \theta \\\ \Rightarrow{{u}_{y}} =16\sin {{60}^{\circ }}\\\ \Rightarrow{{u}_{y}} =16\times \dfrac{\sqrt{3}}{2}\\\ \Rightarrow{{u}_{y}} =8\sqrt{3}m{{s}^{-1}}$.
Let us first find the time taken by water to reach the water. This is equal to the time taken to cover the horizontal distance of $x=8m$. In this case, the water is influenced by the gravitational force exerted by the earth in the downward direction. This means that the projectile will be accelerated downwards with an acceleration equal to the acceleration due to gravity, i.e. g.Since there is no force in the horizontal direction, the horizontal velocity of the projectile will be constant, i.e. ${{u}_{x}}=8m{{s}^{-1}}$.

Therefore, the time taken by the projectile to cover a horizontal distance of $x=8m$ is,
$t=\dfrac{x}{{{u}_{x}}}\\\ \Rightarrow t =\dfrac{8}{8}\\\ \Rightarrow t =1s$.
Now use the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$ for the vertical motion of the projectile to find the vertical position of the projectile when $x=8m$.
In this case, $s=y$, $u={{u}_{y}}=8\sqrt{3}m{{s}^{-1}}$, $t=1s$ and $a=-g=-10m{{s}^{-2}}$ (since the water is being accelerated downwards (-g) in the presence of the gravitational force).
Substitute these values in the above equation.
$\Rightarrow y=(8\sqrt{3})(1)+\dfrac{1}{2}(-10){{1}^{2}}$
$\therefore y=8\sqrt{3}-5=13.85-5=8.85m\approx 8.9m$
This means that the height at which the water strikes the wall 8m away is 8.9m.

Hence, the correct option is A.

Note: This is one of the methods to solve the given question that is first find the time to reach the horizontal position $x=8m$ and then find the vertical position of the projectile at $x=8m$. Another direct way to solve this question is to use the equation of trajectory of a projectile motion. It is used to find one coordinate of a position when the other coordinate is known.