Q. When a projectile is projected at a certain angle with the horizontal, its horizontal range is $R$ and time of flight is $T_{1}$. When the same projectile is throwing with the same speed at some other angle with the horizontal, its horizontal range is $R$ and time of flight is $T_{2}$. The product of $T_{1}$ and $T_{2}$ is
A) $\frac{R}{g}$
B) $\frac{2R}{g}$
C) $\frac{3R}{g}$
D) $\frac{4R}{g}$
Explanation
(b) The two angles of projection are clearly $\theta$ and $(90{^\circ} - \theta)$
$T_{1} = \frac{2v\sin\theta}{g}\ \ \ and\ \ \ T_{2} = \frac{2v\sin{(90{^\circ} - \theta)}}{g}$
$T_{1}T_{2} = \frac{2(v)^{2}(2\sin\theta\cos\theta)}{g \times g} = \frac{2R}{g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.