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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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