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A particle P is projected from a point o...

A particle P is projected from a point on the surface of a smooth inclined pane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same postion. P and Q colide on the inclined plane after `t=4` seconds. Find the speed projection in m/s.

A

`5 m//s`

B

`10 m//s`

C

`15 m//s`

D

`20 m//s`

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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