Q. Two particles $A$ and $B$ are projected with same speed so that the ratio of their maximum heights reached is $3\ :1$. If the speed of $A$ is doubled without altering other parameters, the ratio of the horizontal ranges attained by $A$ and $B$ is
A) $1\ :1$
B) $2\ :1$
C) $4\ :1$
D) $3\ :2$
Explanation
(c) $H = \frac{u^{2}\sin^{2}\theta}{2g}\ \Rightarrow \frac{H_{1}}{H_{2}} = \frac{\sin^{2}\theta_{1}}{\sin^{2}\theta_{2}}$
$\frac{\sqrt{3}}{1} = \frac{\sin\theta_{1}}{\sin\theta_{2}}\ \ So,\ \frac{\cos\theta_{1}}{\cos\theta_{2}} = \frac{1}{\sqrt{3}}$
$\frac{R_{1}}{R_{2}} = \frac{(2u)^{2}\sin{2\theta_{1}}}{u^{2}\sin{2\theta_{2}}} = \frac{4.\sin\theta_{1}\cos\theta_{1}}{\sin\theta_{2}\cos\theta_{2}} = \frac{4}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.