question_answer

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

A)  \[{{30}^{o}}\]            

B)  \[{{45}^{o}}\]

C)  \[{{90}^{o}}\]            

D)  \[{{40}^{o}}\]

Correct Answer: A

Solution :

When body is projected with an initial velocity u, at an angle 6 with the horizontal, then \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] Given, \[R=4\,\sqrt{3}\,H\] \[\therefore \] \[\frac{{{u}^{2}}(2\sin \theta \cos \theta )}{g}=4\sqrt{3}\,\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] \[\Rightarrow \] \[\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\tan \theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\theta ={{30}^{o}}\]. Note: Horizontal range is same whether the body is projected at \[{{30}^{o}}\] or \[({{90}^{o}}-{{30}^{o}}={{60}^{o}})\].