A) \[\sqrt{2/3}\]
B) \[2/\sqrt{3}\]
C) \[\sqrt{3}/2\]
D) \[\sqrt{3}/2\]
Correct Answer: C
Solution :
Let body is projected at an angle 6 with the horizontal with a velocity u. The height attained is given by \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] Given, \[{{H}_{1}}={{H}_{2}}\] \[\therefore \] \[\frac{u_{1}^{2}{{\sin }^{2}}{{\theta }_{1}}}{2\,g}=\frac{u_{2}^{2}\,{{\sin }^{2}}{{\theta }_{2}}}{2\,g}\] \[\Rightarrow \] \[\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{\sin \,{{\theta }_{2}}}{\sin {{\theta }_{1}}}\] \[\Rightarrow \] \[\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{\sin {{60}^{o}}}{\sin {{45}^{o}}}\] \[\Rightarrow \] \[\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{\sqrt{3}/2}{1/\sqrt{2}}\] \[\Rightarrow \] \[\frac{{{u}_{1}}}{{{u}_{2}}}=\sqrt{\frac{3}{2}}\].
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