Question

A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is $\alpha ={30}^o$, and the angle of the barrel to the horizontal $\beta ={60}^o$. The initial velocity $u$ of the shell is $21m/sec$. Then distance of point from the gun at which shell will fall-

• A. $10m$
• B. $20m$
• C. $30m$
• D. $40m$

Answer 1

The correct option is C $30m$

Given,

Initial velocity of shell $u=21m/s$

Slope of the hill $\alpha ={30}^{\circ }$

Angle of the barrel to the horizontal $\beta ={60}^{\circ }$

So,

$\begin{array}{c}\theta =\beta -\alpha \\ ={60}^{\circ }-{30}^{\circ }\\ ={30}^{\circ }\end{array}$

Now,

Distance of the point from the gun at which shell will fall $=$ Maximum range

$\begin{array}{c}R=\frac{2u^2}{g}\frac{\sin \theta \cos \left(\theta +\alpha \right)}{{\cos }^2\alpha }\\ =\frac{2\times {21}^2\times \sin {30}^{\circ }\cos {60}^{\circ }}{9.8\times {\cos }^2{30}^{\circ }}\\ =\frac{2\times 21\times 21\times \frac{1}{2}\times \frac{1}{2}}{9.8\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}}\\ =30m\end{array}$

Hence, Option $C$ is the correct answer.