From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is doubleof what it was at a height h above A. Show that the greatest height attained by the stone is `5/3 h.`

Options

A. `(3)/(5) h`
B. `(5)/(3) h`
C. `(7)/(5) h`
D. `(5)/(7) h`

Answer

B

Explanation

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`