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 $v$ of the shell is $21m{s}^1$. Then find the distance of point from the gun at which the shell will fall.

• A. 30 m
  
• B. 40 m
  
• C. 60 m
  
• D. 80 m
  

Answer 1

The correct option is A 30 m

Here $u=21m{s}^{-1},\alpha ={30}^o,\theta =\beta -\alpha ={60}^o-{30}^o={30}^o$

Maximum range, $R=\frac{2u^2\sin \theta \cos (\theta +\alpha )}{g{\cos }^2}$

$=\frac{2\times (21{)}^2\times \sin {30}^o\cos {60}^o}{9.8\times {\cos }^2{30}^o}=30m$