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The speed at the maximum height of a pro...

The speed at the maximum height of a projectile is `(1)/(3)` of its initial speed u. If its range on the horizontal plane is `(n sqrt(2)u^(2))/(9g)` then value of n is

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The correct Answer is:
4
The speed ………….
At maximum height `v = u cos theta`
`(u)/(3)=u cos theta implies cos theta = (1)/(3)`
`R = (u^(2)sin2theta)/(g) = (2u^(2)sin theta cos theta)/(g)`
`= (2u^(2))/(g)xx(2sqrt(2))/(3)xx(1)/(3)= (4sqrt(2)u^(2))/(9g)`
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