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

To complete a smooth vertical track of radius R, the minimum height at which a particle starts, must be equal to `5/2R` (one can proved it from energy conservation). Thus in our problem body could not reach the upper most point of the vertical track of radius `R//2`. Let the particle A leave the track at some point O with speed v (figure). Now from energy conservation for the body A in the field of gravity:
`mg[h-h/2(1+sin theta)]=1/2mv^2`
or, `v^2=gh(1-sin theta)` (1)
From Newton's second law for the particle at the point `underset(.)O`, `F_n=mw_n`,
`N+mgsin theta=(mv^2)/((h//2))`
But, at the point O the normal reaction `N=0`
So, `v^2=(gh)/(2)sin theta` (2)
From (3) and (4), `sin theta=2/3` and `v=sqrt((gh)/(3))`
After leaving the track at O, the particle A comes in air and further goes up and at maximum height of it's trajectory in air, it's velocity (say `vecv`) becomes horizontal (figure). Hence, the sought velocity of A at this point.
`v^'=vcos(90-theta)=vsin theta=2/3sqrt((gh)/(3))`