When a particle is projected at an angle to the horizontal, it has range R and time of flight `t_(1)`. If the same projectile is projected with same speed at another angle to have the saem range, time of flight is `t_(2)`. Show that:
`t_(1)t_(2)=(2R//g)`

Explanation

`t_(1)=(2u sin theta)/(g)` and `t_(2)=(2u sin (90^(@)-theta))/(g)=(2u cos theta)/(g)`
`therefore t_(1)t_(2)=(2u sin theta)/(g)xx(2u cos theta)/(g)=(2)/(g)((u^(2)sin 2theta)/(g))=(2R)/(g)`