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The trajectory of a projectile in a vert...

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`

Radius of curvature of the path of the projectile at the topmost point P is :-

A

`1/2 m`

B

`1m`

C

`4m`

D

`1/4 m`

Text Solution

Verified by Experts

The correct Answer is:
D
At the top most point `v=u cos theta=sqrt(10) cos 60^(@)=sqrt(10)/2`

`:. Mg=(mv^(2))/R, R=((sqrt(10)/2)^(2))/10=10/40 R=1/4 m`
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Knowledge Check

  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Range OA is :-

    A
    `sqrt(3)/2`
    B
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    C
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    D
    `3/8`

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    `8/3`
    B
    `3/8`
    C
    `sqrt(3)`
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  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-

    A
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    B
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    C
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    D
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