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The kinetic energy of a projectile at th...

The kinetic energy of a projectile at the highest point of its path is found to be `3//4^(th)` of its initial kinetic energy. If the body is projected from the ground, the angle of projection is

A

`0^(@)`

B

`30^(@)`

C

`60^(@)`

D

`40^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K.E_("top") = (3)/(4)(K.E_(i)),(1)/(2)m(u cos theta)^(2) = (3)/(4) ((1)/(2)m u^(2))`
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Knowledge Check

  • The kinetic energy of a projectile at the highest point is-

    A
    Zero
    B
    Maximum
    C
    Minimum
    D
    Equal to total energy

  • The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. The angle of projection with the horizontal is

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

  • At the highest point of the path of a projectile, its

    A
    speed is zero
    B
    spoeed is minimum
    C
    kinetic energy is minimum
    D
    Potential energy is maximum

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