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

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)/2`

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
`t=sqrt((2h)/g)` and `R=usqrt((2h)/g)`
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