A particle is projected horizontally as shown from the rim of a large hemispherical bowl. The displacement of the particle when it strikes the bowl the first time is R. Find the velocity of the particle at that instant and the time taken.

Answer

`uhat(i)+sqrt(sqrt(3)Rg) hat(j), sqrt((sqrt(3)R)/g)`

Explanation

`:.` Vertical displacement of particle`=(Rsqrt(3))/2`

Time for this `=sqrt((2xxRsqrt(3)/2)/g)=sqrt((sqrt(3)R)/g)`
`vec(v)(t)=uhat(i)+g that(j)=uhat(i)+gxxsqrt((sqrt(3)R)/g)hat(j)=uhat(i)+sqrt(sqrt(3)Rg) hat(j)`