A small body A starts sliding from the height h down an inclined groove passing into a half-circle of radius `h//2` (figure).

Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory (after breaking off the groove).

Explanation

When the ball reaches at `P`, its speed

`u=sqrt(2gh)`
Radius of circle `r=(h)/(2)`
`sqrt(2gr)=sqrtgh`
`sqrt(5gr)=sqrt(2.5gh)`
`sqrt(2gr)ltu lt sqrt(5gr)`
That is the body will leave circular path somewhere between `B` and `H`
At `Q:`
`v^(2)=u^(2)-2gr(1+costheta)=gh-ghcostheta (i)`
`N+mgcostheta=(mv^(2))/(h//2)` (ii)
When the body loses contact with surface.
`N=0` (iiI)
From (i), (ii) and (iii) we get
`(gh)/(2)costheta=gh-ghcosthetarArrcostheta=(2)/(3)`
`v=sqrt((gh)/(2)costheta)=sqrt((gh)/(3))`
At `Q`, the body losses contact, now the body moves as a free particle in a parabolic path (projectile motion). In a projectile motion, the velocity at the highest point is horizontal and equal to `vcostheta`.
`vcostheta=(2)/(3)sqrt((gh)/(3))`