Q. For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_{1}$ and $t_{2}$ are the times of flight in the two cases then
A) $t_{1}t_{2} \propto R^{2}$
B) $t_{1}t_{2} \propto R$
C) $t_{1}t_{2} \propto \frac{1}{R}$
D) $t_{1}t_{2} \propto \frac{1}{R^{2}}$
Explanation
(b) For same range angle of projection should be $\theta$ and $90 - \theta$
So, time of flights $t_{1} = \frac{2u\sin\theta}{g}$ and
$t_{2} = \frac{2u\sin{(90 - \theta)}}{g} = \frac{2u\cos\theta}{g}$
By multiplying $= t_{1}t_{2} = \frac{4u^{2}\sin\theta\cos\theta}{g^{2}}$
$t_{1}t_{2} = \frac{2}{g}\frac{(u^{2}\sin{2\theta})}{g} = \frac{2R}{g} \Rightarrow t_{1}t_{2} \propto R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.