question_answer

For a projectile\[{{(range)}^{2}}\]is 48 times of \[{{(maximum\text{ }height)}^{2}}\]obtained. Find the angle of projection.

A)  \[60{}^\circ \]           

B)  \[30{}^\circ \]

C)  \[45{}^\circ \]             

D)  \[75{}^\circ \]

Correct Answer: B

Solution :

 As shown in figure, distance OB is the range R. Maximum height of projectile \[H=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}\] Now, it is given that \[{{(Range)}^{2}}=48{{(maximum\text{ }height)}^{2}}\] \[\therefore \] \[{{\left( \frac{{{u}^{2}}\sin 2\alpha }{g} \right)}^{2}}=48{{\left( \frac{{{u}^{2}}{{\sin }^{2}}2\alpha }{2g} \right)}^{2}}\] Or \[\frac{{{u}^{2}}\sin 2\alpha }{g}=4\sqrt{3}\left( \frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g} \right)\] Or \[\frac{2\sin \alpha \cos \alpha }{g}=\frac{{{\sin }^{2}}\alpha }{2}\] Or \[\tan \alpha =\frac{4}{4\sqrt{3}}=\frac{1}{\sqrt{3}}\] \[\therefore \] \[\alpha =30{}^\circ \]