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)

$R$

Answer 2

A .projectile can have same range if angles of projection are complementary i.e., 0 and $(90^? - \theta)$. Thus, in both cases : $t_{1}=\frac{2u\,sin\,\theta}{g}\,...\left(i\right)$ $t_{2}=\frac{2u\,sin\,\left(90^{?}-\theta\right) }{g}$ $=\frac{2u\,cos\,\theta}{g}\,...\left(ii\right)$ From Eqs. $\left(i\right)$ and $\left(ii\right)$ $t_{1}t_{2}=\frac{4u^{2}\,sin\,\theta cos\,\theta }{g^{2}}$ $t_{1}t_{2}=\frac{2u^{2}\,sin\,2\theta }{g^{2}}$ $=\frac{2}{g} \frac{u^{2}\,sin\,2\theta}{g}$ $\therefore t_{1}t_{2}=\frac{2R}{g}\,\left(\therefore R=\frac{u^{2}\,sin\,2\theta}{g}\right)$ Hence, $t_{1}t_{2}\,\propto\,R$