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The velocity at the maximum height of a ...

The velocity at the maximum height of a projectile is half of its velocity of projection `u`. Its range on the horizontal plane is

A

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

B

`(sqrt(3)u^(2))/(2g)`

C

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

D

`(u^(2))/(2g)`

Text Solution

Verified by Experts

The correct Answer is:
B
At maximum height `v=u cos theta`
`(u)/(2)=vrArrcos theta=(1)/(2)rArrtheta=60^(@)`
`R=(u^(2)sin2theta)/(g)=(u^(2)sin(120^(@)))/(g)`
`=(u^(2)cos30^(@))/(g)=(sqrt(3)u^(2))/(2g)`
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