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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

`R=2H` given
We know `R=4Hcotthetaimpliescottheta=(1)/(2)`
From triangle we can say that `sintheta=(2)/(sqrt5)`,`costheta=(1)/(sqrt5)`
Range of projectile `R=(2v^2sinthetacostheta)/(g)`
`=(2v^2)/(g)xx(2)/(sqrt5)xx(1)/(sqrt5)=(4v^2)/(5g)`
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