From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is

Options

A. `a) (3)/(5)h`
B. b)`(5)/(3) h`
C. c)`(7)/(5) h`
D. d)`(5)/(7) h`

Answer

B

Explanation

From equations of motions, we know
`v^(2) = u^(2) - 2gh" "...(i)`
Here, `v=2v`, so
`(2v)^2 = u^(2) + 2gh" "...(ii) `
Now, on adding Eqs. (i) and (ii), we get
`5v^(2) + 2u^(2)`
`u^(2) = (5)/(2) v^(2)" "...(iii)`
On subtracting Eq. (i) in Eq. (ii), we get
`3v^(2) = 4gh`
`v^(2) = (4)/(3) gh" "....(iv)`
And we know that the
`H= (u^2)/( 2g)`
So, here `H = ((5)/(2) v^(2) )/( 2g) = ((5)/(2) ((4)/(3) gh) )/( 2g)`
`H= (5)/(3) h`