A) 3 : 1
B) 1 : 3
C) 1 : 2
D) 2 : 1
Correct Answer: B
Solution :
As\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]\ \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{\sin {{\theta }_{2}}}=\frac{{{\sin }^{2}}30{}^\circ }{{{\sin }^{2}}60}\]=\[\frac{1/4}{3/4}=\frac{1}{3}\]You need to login to perform this action.
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