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(D) Let the initial velocity of the particle be $u$. The initial kinetic energy is $K = \frac{1}{2}mu^2$.At the maximum height,the vertical component

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$K/4$

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(D) Let the initial velocity of the particle be $u$. The initial kinetic energy is $K = \frac{1}{2}mu^2$.At the maximum height,the vertical component of velocity becomes $0$,but the horizontal component remains constant.The horizontal component of velocity is $v_x = u \cos 	heta$.Given $	heta = 60^o$,so $v_x = u \cos 60^o = u/2$.The kinetic energy at the maximum height is $K' = \frac{1}{2}m(v_x)^2$.Substituting $v_x = u/2$,we get $K' = \frac{1}{2}m(u/2)^2 = \frac{1}{2}m(u^2/4) = \frac{1}{4} (\frac{1}{2}mu^2)$.Since $K = \frac{1}{2}mu^2$,we have $K' = K/4$.

$A$ particle is projected with kinetic energy $K$ at an angle of $60^o$ with the horizontal. The kinetic energy of the particle at the maximum height is . . . . . . .

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