A) \[\sqrt{2/3}\]
B) \[2/\sqrt{3}\]
C) \[\sqrt{3/2}\]
D) \[\sqrt{3}/2\]
Correct Answer: C
Solution :
Vertical height\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Since, vertical height is same \[\therefore \] \[\frac{u_{1}^{2}{{\sin }^{2}}{{\theta }_{1}}}{2g}=\frac{u_{2}^{2}{{\sin }^{2}}{{\theta }_{2}}}{2g}\] \[\Rightarrow \] \[\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{\sin {{\theta }_{2}}}{\sin {{\theta }_{1}}}\] \[=\frac{\sin {{60}^{o}}}{\sin {{45}^{o}}}\] \[=\frac{\sqrt{3}/2}{1/\sqrt{2}}\] \[=\sqrt{\frac{3}{2}}\]
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