question_answer

The horizontal range of a projectile is \[4\sqrt{3}\] times the maximum height achieved by it, then the angle of projection is:

A) \[30{}^\circ \]   

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

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

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

Correct Answer: A

Solution :

                 Let body projected with initial velocity u at an angle. with the horizontal. Then                 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\]                 \[{{H}_{m}}=\frac{{{u}^{2}}\sin \theta }{2g}\] Given,   \[\frac{R}{{{H}_{m}}}=4\sqrt{3}\]                 \[\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}=\frac{4\sqrt{3}{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \]               \[\cos \theta =\sqrt{3}\sin \theta \] \[\Rightarrow \]               \[\cos \theta =\sqrt{3}\] \[\Rightarrow \]               \[\theta ={{30}^{o}}\]