A particle is projected from horizontal ground at angle `'theta'` with speed 'u'. In same plane of motion a horizontal acceleration 'a' exists so that projected particle returns back to point of projection. Find time of flight.

Options

A. `(2u sin theta)/(g)`
B. `(3usin theta)/(g)`
C. `(2u cos theta)/(alpha)`
D. `(3u cos theta)/(alpha)`

Answer

a,c

Explanation

`T=(2u sin theta)/(g)` or `(2u sin theta)/(a)`