Two particles are projected simultaneously with the same speed `v` in the same vertical plane with angles of elevation `theta`, and `2theta`, where `theta lt 45^(@)`.At what time will velocities be parallel?

Options

A. `t=v/(g) "tan"theta/2 "cosec"(3theta)/2`
B. `t=v/(g) "cos"theta/2 "cot"(3theta)/2`
C. `t=v/(g)"cos"theta/2 "tan"(3theta)/2`
D. `t=v/(g) "cos" theta/2 "cosec"(3theta)/2`

Answer

D

Explanation

Velocity of particle after time `t` is `vecV_(1)=(v cos theta hati+v sin theta hatj)-(ghatj)t`
`vecV_(2)=(v cos 2theta hati+v sin 2theta hatj)-(ghatj)t`
To be parallel of `V_(1)` and `V_(2)`
`rArr (v cos theta)/(v cos 2theta)=(v sin theta-"gt")/(v sin 2theta-"gt")`
Solving the above equation, we get,
`t=v/gcos(theta/2)cosec(3theta)/2`