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

Answer 1

Answer: (D)

A projectile can have same range if angles of projection are complementary ie, $\theta$ and $\left(90^{\circ}-\theta\right)$.

In both cases
$t_{1}=\frac{2 u \sin \theta}{g}$ ... (i)
$t_{2}=\frac{2 u \sin \left(90^{\circ}-\theta\right)}{g}$
$=\frac{2 u \cos \theta}{g}$ ... (ii)
From Eqs. (i) and (ii)
$t_{1} t_{2}=\frac{4 u^{2} \sin \theta \cos \theta}{g^{2}}$
$t_{1} t_{2} =\frac{2 u^{2} \sin 2 \theta}{g^{2}} $
$=\frac{2}{g} \frac{u^{2} \sin 2 \theta}{g} $
$\therefore t_{1} t_{2} =\frac{2 R}{g}$
$ \left(\therefore R=\frac{u^{2} \sin 2 \theta}{g}\right)$
Hence, $ t_{1} t_{2} \propto R$