A body of mass M kg is on the .top point of a smooth hemisphere of radius 5 m. It is released to slide down the surface of the hemisphere. It leaves the surface when velocity is `5 ms^-1`, At this instant the angle made by the radius vector of the body with the vertical is (acceleration due to gravity =`10 ms^-2`)

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

Comparing ,`x=6t` with `x=(ucostheta)`and `y=48t-4.9t^2`with`y=(usintheta)t-1/2g t^2` then,
`impliesucostheta=36"and"u=sintheta=48`
Squaring and adding on the sides,
`u^2cos^2theta+u^2sin^2theta =(36)^2+(48)^2`
`u^2=(36)^2+(48)^2`
`:.u=60m//s`