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, \[{{\theta }_{1}}=\frac{\pi }{3},{{\theta }_{2}}=\frac{\pi }{6},{{H}_{1}}=Y\] \[\therefore \] \[{{H}_{2}}=\frac{{{\sin }^{2}}{{\theta }_{2}}}{{{\sin }^{2}}{{\theta }_{1}}}{{H}_{1}}\] \[{{H}_{2}}=\frac{{{\sin }^{2}}\frac{\pi }{6}}{{{\sin }^{2}}\frac{\pi }{3}}Y\] \[{{H}_{2}}=\frac{Y}{3}\]You need to login to perform this action.
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