question_answer

The horizontal range of projectile is \[4\sqrt{3}\]times of its maximum height. The angle of projection will be:

A)  \[{{40}^{o}}\]            

B)  \[{{90}^{o}}\]

C)  \[{{30}^{o}}\]            

D)  \[{{45}^{o}}\]

Correct Answer: C

Solution :

 Let a body be projected with initial velocity u, at an angle \[\theta \] with the horizontal, then the horizontal range is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] and \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] Given, \[R=4\sqrt{3}\,H\] Using \[\sin 2\theta =2\sin \theta \cos \theta \], we have \[\frac{{{u}^{2}}\times 2\sin \theta \cos \theta }{g}=4\sqrt{3}\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] \[\Rightarrow \] \[\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\tan \theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\theta ={{30}^{o}}\] Note: Range is same whether projected at \[\theta \] or \[{{90}^{o}}-\theta \].