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A particle is dropped from a height h. A...

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

A

`d^(2)=(u^(2)h)/(2g)`

B

`d^(2)=(2u^(2)h)/g`

C

`d=h`

D

`gd^(2)=u^(2)h`

Text Solution

Verified by Experts

The correct Answer is:
B
Time to reach at ground`=sqrt((2h)/g)`

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