Q. A projectile can have the same range $R$ for two angles of projection. If $T_{1}$ and ${\ T}_{2}\ $be the times of flights in the two cases, then the product of the two times of flights is directly proportional to
A) $\frac{1}{R^{2}}$
B) $\frac{1}{R}$
C) $R$
D) $R^{2}$
Explanation
(c) We know that range of projectile is same for complementary angles $ie,$ for $\theta\ and\ (90{^\circ} - \theta).$
$\therefore\ T_{1} = \frac{2u\sin\theta}{g}$
$T_{2} = \frac{2u\sin(90{^\circ} - \theta)}{g} = \frac{2u\cos\theta}{g}$
$and\ \ \ R = \frac{u^{2}\sin\theta}{g}$
$Therefore,\ \ T_{1}T_{2} = \frac{2u\sin\theta}{g} \times \frac{2u\cos\theta}{g}$
$= \frac{{2u}^{2}(2\sin{\theta\cos{\theta)}}}{g^{2}}$
$= \frac{{2u}^{2}(2\sin{\theta)}}{g^{2}} = \frac{2R}{g}$
$\therefore\ T_{1}T_{2} \propto R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.