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

Answer

B

Explanation

Time to reach at ground`=sqrt((2h)/g)`

In this time horizontal displacement
`d=uxxsqrt((2h)/g)rArr d^(2)=(u^(2)xx2h)/g`