A particle is dropped from a height h. Another particle which is initially at a horizontal distance d from the first is simultaneously projected with a horizontal velocity u and the two particles just collide on the ground. Then

Options

A. `d^2 = (u^2h)/(2h)`
B. `d^2 = (2u^2h)/g`
C. `d=h`
D. `gd^2 = u^2h`

Answer

B

Explanation

`T = (sqrt (2h)/g)`
`d = (u)T = u (sqrt(2h)/g)`
`:. d^2 = (2hu^2)/g` .