Question

A projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the times of flight in the two cases, then $R$ is

• A. $T_1{T}_{2}g$
• B. $\frac{T_1{T}_{2}g}{2}$
• C. $(T_1^2+T_1^2)g$
• D. $\frac{T_1^2+T_2^2}{2}g$

Answer 1

The correct option is B $\frac{T_1{T}_{2}g}{2}$

We know that for the two different angle of projection with the same range, the condition is the angles of projection should be complementary.
Let $\left(\alpha \right)$ and $(90-\alpha )$ be angle of projection, measured from the horizontal.

We know that ,time of flight is $T=\frac{2u\sin \theta }{g}$, here $\theta$ is angle of projection.

For the two different angle with same range

$T_1=\frac{2u\sin \alpha }{g},T_2=\frac{2u\sin (90-\alpha )}{g}=\frac{2u\cos \alpha }{g}$

Range of projectile:
$R=\frac{u^2\sin 2\alpha }{g}=\frac{2u^2\sin \alpha \cos \alpha }{g}$
$\Rightarrow R=\frac{T_1{T}_{2}g}{2}$

Hence B is the correct option.