A projectile is projected with some initial velocity and some initial angle of projection. A wind is also blowing, due to which constant horizontal retardation a is imparted to the particle in the plane of motion. It is found that, the particle is at same height at two different time `t_(1)` & `t_(2)` and particle is at same horizontal distance at two different time `t_(3)` & `t_(4)`
Angle of projection of particle is :-

Options

A. `tan^(-1)((t_(1)+t_(2))/(t_(3)+t_(4)))`
B. `tan^(-1)[(a(t_(1)+t_(2)))/(g(t_(3)+t_(4)))]`
C. `tan^(-1)[(g(t_(1)+t_(2)))/(a(t_(3)+t_(4)))]`
D. None of these

Answer

C

Explanation

In vertical direction `h=(u sin theta)t-1/2 g t^(2)`
`rArr t^(2)-((2u sin theta)/g)t+(2h)/g=0`
`rArr t_(1)+t_(2)=(2u sin theta)/g` …(i)
In horizontal direction `x=(u cos theta)t-1/2 at^(2)`
`rArr t^(2)-((2u cos theta)/a)t+(2x)/a=0`
`rArr t_(3)+t_(4)=(2u cos theta)/a`...(ii)
From (i) and (ii) `theta=tan^(-1)[(g(t_(1)+t_(2)))/(a(t_(3)+t_(4)))]`