The distance x of a particle moving in one dimensions, under the action of a constant force is related to time t by the equation, `t=sqrt(x)+3`, where x is in metres and t in seconds. Find the displacement of the particle when its velocity is zero.

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`