Question

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 flight is proportional to:

• A. $R^2$
• B. $\frac{1}{R^2}$
• C. $\frac{1}{R}$
• D. $R$

Answer 1

The correct option is D $R$

$t_1=\frac{2v\sin \theta }{g}$

$t_2=\frac{2v\sin \left({90}^0-\theta \right)}{g}=\frac{2v\cos \theta }{g}$

Now$,$ $t_1{t}_2=\frac{\left(2v\sin \theta \right)\left(2v\cos \theta \right)}{g^2}$

$\Rightarrow \frac{2}{g}\left[\frac{v^2\left(2\sin \theta \cos \theta \right)}{g}\right]=\frac{2}{g}\frac{v^2\sin 2\theta }{g}=\frac{2}{g}=R$

$\therefore t_1{t}_2\alpha R$

Hence,

option $\left(D\right)$ is correct answer.