question_answer

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

A)  \[{{45}^{o}}\]          

B)  \[{{60}^{o}}\]

C)  \[{{90}^{o}}\]          

D)  \[{{30}^{o}}\]

Correct Answer: D

Solution :

Let u be initial velocity of projection at angle \[\theta \]with the horizontal. Then, horizontal range, \[R=\frac{{{u}^{2}}\,\,\sin \,\,2\theta }{g}\] and maximum height, \[H=\frac{{{u}^{2}}\,\,{{\sin }^{2}}\,\,\theta }{2g}\] Given.       \[R=4\sqrt{3}H\] \[\therefore \] \[\frac{{{u}^{2}}\,\sin \,2\theta }{g}=4\sqrt{3}.\frac{{{u}^{2}}\,{{\sin }^{2}}\,\theta }{2g}\] \[\therefore \] \[2\sin \theta \,\cos \theta =2\sqrt{3}{{\sin }^{2}}\theta \] or \[\frac{\cos \,\theta }{\sin \theta }=\sqrt{3}\] or \[\cot \theta =\sqrt{3}=\cot \,{{30}^{o}}.\] \[\Rightarrow \] \[\theta ={{30}^{o}}\]