Question

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42. If a particle is projected 60° to the horizontal with a kineti
energy E. The kinetic energy at the highest point will be
(a) / (6) E (c) zero (d) -
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Answer 1

Answer:

Explanation:

let u be the initial velocity,

∴ initial kinetic energy is K = (1/2)mu²,

at the maximum height velocity is u cos60 = (u/2),

∴ kinetic energy at the maximum height is KE=(1/2)m(u/2)²,

                                                                       =(1/4)k is the answer,

You gave wrong options.

Hope it helps.