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The total speed of a projectile at its g...

The total speed of a projectile at its greater height is `sqrt(6/7)` of its speed when it is at half of its greatest height. The angle of projection will be :-

A

`60^(@)`

B

`45^(@)`

C

`30^(@)`

D

`50^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`v_(y)^(2)=u^(2)sin^(2)theta-2gxxH/2,v_(x)^(2)=u^(2) cos^(2) theta`

`:. U cos theta =sqrt(6/7)[sqrt(v_(x)^(2)+v_(y)^(2))]rArr cos theta=sqrt(3)/2` or `theta=30^(@)`
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Knowledge Check

  • 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
    `60^(@)`
    C
    `45^(@)`
    D
    `tan^(-1)(3//4)`

  • The velocity of a projectile when it is at the greatest height is (sqrt (2//5)) times its velocity when it is at half of its greatest height. Determine its angle of projection.

    A
    `theta=30^@`
    B
    `theta=45^@`
    C
    `theta=60^@`
    D
    `theta=90^@`

  • The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

    A
    `sin^(1) x`
    B
    `cos^(-1)x`
    C
    `sin^(-1) (1//x)`
    D
    `cos^(-1) (1//x)`

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