question_answer

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

A) \[45{}^\circ \]                                   

B) \[60{}^\circ \]

C) \[90{}^\circ \]

D) \[30{}^\circ \]

Correct Answer: D

Solution :

                Let u be initial velocity of projection at angle 9 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           \[\cot \theta =\sqrt{3}=\cot {{30}^{o}}\]