Question

A projectile can have the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be times of flight in two cases, then the product of the two times of flight is directly proportional to

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

Answer 1

The correct option is A $R$

We know that projectile has the same range for two complementary angles $\theta$ and ${90}^{\circ }-\theta$.


So the times of flight for these motion will be,

$T_1=\frac{2u\sin \theta }{g}\&T_2=\frac{2u\cos \theta }{g}$

Then,
$T_1{T}_2=\frac{4u^2\sin \theta \cos \theta }{g}=2R$

$(\because R=\frac{u^2\sin 2\theta }{g})$

Thus, it is proportional to $R($Range$).$

Hence, option $\left(\text{A}\right)$ is correct.