Question

When a particle is projected at an angle to the horizontal, it has range R and time of flight t1. If the same projectile is projected with the same speed at another angle to have the same range and time of flight $t_2$ then,

• A. $t_1+t_2=\frac{2R}{g}$
• B. $t_1,t_2=\frac{2R}{g}$
• C. $t_1-t_2=\frac{2R}{g}$
• D. $t_1{t}_2=\frac{2R}{g}$

Answer 1

The correct option is D

$t_1{t}_2=\frac{2R}{g}$

Let $\theta$ and ${90}^{\circ }-\theta$ be the angles of projection for the same range
$t_1=\frac{2usin\theta }{g}and{t}_2=2usin\frac{[{90}^{\circ }-\theta ]}{g}=\frac{2ucos\theta }{g}{t}_1{t}_2=\frac{4u^{2}sin\theta cos\theta }{g}=\frac{2R}{g}\therefore [R=\frac{2u^{2}sin\theta cos\theta }{g}]$