A particle is projected horizontally from a tower with velocity `V_(0)`. Wind is blowing in opposite direction and is providing a constant horizontal acceleration `a_(0)`. If particle strikes the ground vertically then height of building is

Options

A. `(V_(0)^(2)g)/(a_(0)^(2))`
B. `(V_(0)^(2)g)/(2a_(0)^(2))`
C. `(2V_(0)^(2)g)/(a_(0)^(2))`
D. `(V_(0)^(2))/(g)`

Answer

B

Explanation

`(1)/(2)g t^(2) = h rArr t = sqrt((2h)/(g))`
`v_(0) - a_(0)t = 0 rArr (v_(0)^(2))/(a_(0)) = (2h)/(g)`
`rArr h = (v_(0)^(2) g)/(2a_(0)^(2))`