A) \[1:3\]
B) \[2:1\]
C) \[3:1\]
D) \[1:4\]
Correct Answer: A
Solution :
If body is projected with initial velocity u at an angle \[\theta \] with the horizontal, then the height H reached by projectile is given by \[H=\frac{{{u}^{2}}\,{{\sin }^{2}}\,\theta \,}{2\,g}\] where g is acceleration due to gravity. Given, \[{{\theta }_{1}} = 30{}^\circ and {{\theta }_{2}} = 60{}^\circ \] \[\therefore \,\,\,\,\,\,\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}=\frac{{{\sin }^{2}}\,(30{}^\circ )}{{{\sin }^{2}}\,(60{}^\circ )}=\frac{{{\left( \frac{1}{2} \right)}^{2}}}{{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=\frac{1}{3}\]You need to login to perform this action.
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