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 flights is proportional to

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

Answer 1

The correct option is C $R$

Time of flight of projectile
$T=\frac{2u\sin \theta }{g}$
Hence $t_1{t}_2=\frac{2u\sin \theta }{g}\times \frac{2u\sin \alpha }{g}......\left(i\right)$
Since in both case range is same, $\alpha$ and $\theta$ is complementary angle.
$\alpha +\theta ={90}^o$
From eqn. $\left(i\right)$
$t_1{t}_2=\frac{2u\sin \theta }{g}\times \frac{2u\sin ({90}^o-\theta )}{g}$
$t_1{t}_2=\frac{2u\sin \theta \times 2u\cos \theta }{g^2}$
$t_1{t}_2=\frac{2}{g}\times \frac{u^2\sin 2\theta }{g}$
$t_1{t}_2=\frac{2R}{g}$
$t_1{t}_2\propto R$