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A particle is projected with a velocity ...

A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is

A

`2v^(2)//3g`

B

`v^(2)//2g`

C

`v^(2)//g`

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
R = 2H
`sin2theta=sin^(2)theta`
`2sintheacostheta=sin^(2)theta`
`:.tantheta=2`
Range `R=(u^(2).2sinthetacostheta)/g`
`=u^(2).2xx(2x)/(sqrt(5)x).x/(sqrt(5)x)xx1/g`
`(u^(2).4)/(5g)`
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