A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height reached by the ball above the top of tower.

Explanation

If AB is the tower then according to the problem velocity at P is given as twive the velocity at Q.
`V_(P)=sqrt((u^(2+2gh)),V_(Q))=sqrt((u^(2)-2gh)),V_(P)=2(V_(Q))`
`sqrt((u^(2)+2gh))=2sqrt((u^(2)+2gh))impliesu^(2)=(10gh)/(3)`
From the top of the tower maximum height reached
`H=(u^(2))/(2g)impliesH=(((10gh)/(3)))/(2g)=(5h)/(3)`