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A particle is projected with a velocity ...

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
Given `R=2H, Tan theta =2, R=(u^(2) 2 sin theta cos theta)/(g)`
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