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The speed of a projectile when it is at ...

The speed of a projectile when it is at its greatest height is `sqrt(2//5)` times its speed at half the maximum height. The angle of projection is

A

`30^(@)`

B

`45^(@)`

C

`60^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
speed at maximum height is `u cos theta` speed of half of the maximum height is
`v=sqrt(h^(2)cos^(2)+theta+v_(y)^(2)`
where `v_(y)^(2)=u^(2) sin^(2) theta-2g (H)/(2)`
`v_(y)^(2)=(u^(2) sin^(2) theta)/(2)`
`3u^(2) cos^(2) theta=u^(2) sin^(2)theta`
`tan theta= sqrt(3)`
`theta=60^(@)`
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