A cricket ball is hit at 300 to the horizontal with a kinetic energy, K. The kinetic energy at the highest point is:A] KB](K root3)/2C] K/2D] 3K/4

Let, u be the initial velocity. So, initial kinetic energy is, K = (1/2)mu2

At the maximum height, velocity is u cos30 = (√3×u/2)

Therefore, kinetic energy at the maximum height is, KE f = (1/2)m(√3×u/2) 2 = (3/4)K

hence option (D) is correct.

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