question_answer

Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \[\frac{\pi }{3}\] and its maximum height is Y, the maximum height attained by the other stone with \[\frac{\pi }{6}\] angle of projection is :

A)  Y                                            

B)  2Y

C)  3Y                                         

D) \[\frac{\pi }{6}\]

Correct Answer: D

Solution :

                Let u be velocity of projection at angle\[\frac{\pi }{3}\]and \[\frac{\pi }{6}\]with the horizontal. Then                 \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]                 \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\] Given, \[{{\theta }_{1}}=\frac{\pi }{3},{{\theta }_{2}}=\frac{\pi }{6},{{H}_{1}}=Y\] \[\therefore \]  \[{{H}_{2}}=\frac{{{\sin }^{2}}{{\theta }_{2}}}{{{\sin }^{2}}{{\theta }_{1}}}{{H}_{1}}\]                 \[{{H}_{2}}=\frac{{{\sin }^{2}}\frac{\pi }{6}}{{{\sin }^{2}}\frac{\pi }{3}}Y\]                 \[{{H}_{2}}=\frac{Y}{3}\]