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A particle is projected with a velocity ...

A particle is projected with a velocity `v` so that its range on a horizontal plane is twice the greatest height attained. If `g` is acceleration due to gravity, then its range is

A

`(4v^(2))/(5g)`

B

`(4g)/(5v^(2))`

C

`(v^(2))/g`

D

`(4v^(2))/sqrt5g`

Text Solution

Verified by Experts

The correct Answer is:
A
Given `R=2H,Tantheta=2,R=(u^(2)2SinthetaCostheta)/g`
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