Correct Answer - B
`v^(2)` (at end of track) `= 2gh`
Let body break at angle `theta` the `(m u^(2))/((h)/(2)) = mg cos theta`…(1)
`u^(2)=v^(2)-2g "h/(2)(1+cos theta)`
solving `cos theta = (2)/(3)` & `u = sqrt((2)/(3)gh)`
`v` at highest `Pt` is
`u cos theta = sqrt((2)/(3)) gh xx (2)/(3) = sqrt((8)/(27)gh)`.