Q. Two stones thrown at different angles have same initial velocity and same range. If $H$ is the maximum height attained by one stone thrown at an angle of $30{^\circ}$, then the maximum height attained by the other stone is
A) $\frac{H}{2}$
B) $H$
C) $2H$
D) $3H$
Explanation
(d) Since range is given to be the same therefore the other angle is $(90{^\circ} - 30{^\circ}),\ ie,\ 60{^\circ}$
$H = \frac{v^{2}\sin^{2}{30{^\circ}}}{2g} = \frac{1}{4}\left\lbrack \frac{v^{2}}{2g} \right\rbrack$
$H^{'} = \frac{v^{2}\sin^{2}{60{^\circ}}}{2g} = \frac{3}{4}\left\lbrack \frac{v^{2}}{2g} \right\rbrack$
$\frac{H^{'}}{H} = \frac{3}{4} \times \frac{4}{1} = 3\ \ or\ \ \ H^{'} = 3H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.