A particle is dropped from point P at time t = 0. At the same time another particle is thrown from point O as shown in the figure and it collides with the particle P. Acceleration due to gravity is along the negative y-axis. If the two particles collide 2 s after they start, find the initial velocity `v_0` of the particle which was projected from O. Point O is not necessarily on ground.

Options

A. `sqrt(6)m//s^(-1),theta=tan^(-1)(1)` with x-axis
B. `sqrt(26)m//s^(-10theta=tan^(-1)(5)` with x-axis
C. `sqrt(2)m//s^(-1),theta=tan^(-1)(2)` with x-axis
D. `sqrt(13)m//s^(-1),theta=tan^(-1)(4)` with axis

Answer

b

Explanation

`x=(u cos theta)t=2, u cos theta=1`
`(u sin theta)t-(1)/(2)g t^(2)=-10, u sin theta=5`
`v=sqrt((u cos theta)^(2)+(u sin theta)^(2))=sqrt(26), tan theta=5`