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 i.e., 0 and $(90^{°}-\theta )$. Thus, in both cases :
$t_1=\frac{2usin\theta }{g}...\left(i\right)$
$t_2=\frac{2usin\left(90^{°}-\theta \right)}{g}$
$=\frac{2ucos\theta }{g}...\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$
$t_1{t}_2=\frac{4u^{2}sin\theta cos\theta }{g^2}$
$t_1{t}_2=\frac{2u^{2}sin2\theta }{g^2}$
$=\frac{2}{g}\frac{u^{2}sin2\theta }{g}$
$\therefore t_1{t}_2=\frac{2R}{g}\left(\therefore R=\frac{u^{2}sin2\theta }{g}\right)$
Hence, $t_1{t}_2\propto R$