Question

A jet of water is projected at an angle $\theta ={45}^{\circ }$ with the horizontal from a point which is at a distance of $x=15\text{m}$ from a vertical wall as shown in the figure. If the speed of projection is $10\sqrt{2}\text{m/s}$, find out the height from ground at which the water jet strikes vertical wall. Take $g=10{\text{m/s}}^2$

• A. $5\text{m}$
• B. $3.75\text{m}$
• C. $7.5\text{m}$
• D. $2.5\text{m}$

Answer 1

The correct option is B $3.75\text{m}$

From the equation of trajectory for a projectile, projected at an angle $\theta$ from horizontal :
$y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{\cos }^2\theta }.........\left(i\right)$
where;
$y=$displacement in y-direction
$x=$displacement in x-direction
$\theta ={45}^{\circ }$, $g=10{\text{m/s}}^2,u=10\sqrt{2}\text{m/s}$
When the water jet strikes the vertical wall $x=15\text{m}$, putting the value in equation $\left(i\right)$:
$y=15\tan {45}^{\circ }-\frac{1}{2}\times \frac{10\times {15}^2}{(10\sqrt{2}{)}^2\times (\frac{1}{2})}$
$\Rightarrow y=15-11.25$
$\therefore y=3.75\text{m}$
The displacement in y-direction represent the vertical distance of jet from ground.