A) \[1:1\]
B) \[2:1\]
C) \[4:1\]
D) \[3:2\]
Correct Answer: C
Solution :
Let\[{{\theta }_{1}}\,\,~and\,\,{{\theta }_{2}}\]be the angles of projection, of two particles A and B and u be their velocity of projection. As per equations, \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\mu }^{2}}{{\sin }^{2}}{{\theta }_{1}}/2g}{{{\mu }^{2}}{{\sin }^{2}}{{\theta }_{2}}/2g}\] \[=\,\,\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}=\frac{3}{1}\] \[\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}=\sqrt{3}=\tan \,\,60{}^\circ \,\,\Rightarrow \,\frac{\sin \,60{}^\circ }{\cos \,30{}^\circ }\,=\frac{\sin \,60{}^\circ }{\sin \,30{}^\circ }\] \[\therefore \,\,\,\,\,\,\,\,\,{{\theta }_{1}} = 60{}^\circ and {{\theta }_{2}} = 30{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
