A particle P is projected from a point on the surface of smooth inclined plane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide after`t=4`. The speed of projection of P is

Options

A. `5 m//s`
B. `10 m//s`
C. `15 m//s`
D. `20 m//s`

Answer

B

Explanation

Time of flight `4=(2u sin theta)/(g cos 60^(@))` …(i)
(angle of projection `=theta`)
Distance travelled by Q on incline in 4 secs is
`=0+1/2xx(sqrt(3)g)/2xx4^(2)=40sqrt(3)`
& the range of particle 'P' is `40 sqrt(3)`
`=u cos thetaxx4+1/2(sqrt(3)g)/2xx4^(2)=40sqrt(3)`
`=u cos theta=0,` so `theta=90^(@)`
from equation (i) `u=10 m//s`