Correct Answer - B

At height h above P:

`v^(2) = u^(2) - 2gh` ....(i)

At depth h below P:

`(2v)^(2) = u^(2) + 2gh`....(ii)

Adding eqns. (i) and (ii)

`5v^(2) = 2u^(2)`

`u^(2) = (5)/(2) v^(2)`

Eqn. (ii) minus Eqn. (i) gives

`3v^(2) = 4 g h rArr v^(2) = (4)/(3) gh`

The maximum height attained by the stone is

`H = (u^(2))/(2g) = (5v^(2)//2)/(2g) = (5v^(2))/(4g) = (5)/(4g) ((4)/(3) gh) = (5)/(3) h`