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(C) The initial kinetic energy of the particle is given by $K = \frac{1}{2} m v^2$,where $v$ is the initial velocity.At the highest point of the traje

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$\frac{K}{4}$

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(C) The initial kinetic energy of the particle is given by $K = \frac{1}{2} m v^2$,where $v$ is the initial velocity.At the highest point of the trajectory,the vertical component of the velocity becomes zero,and the velocity of the particle is equal to the horizontal component,which is $v_x = v \cos 	heta$.Given $	heta = 60^{\circ}$,the velocity at the highest point is $v_x = v \cos 60^{\circ} = \frac{v}{2}$.The kinetic energy at the highest point $K'$ is given by $K' = \frac{1}{2} m (v_x)^2$.Substituting the value of $v_x$,we get $K' = \frac{1}{2} m (\frac{v}{2})^2 = \frac{1}{2} m (\frac{v^2}{4}) = \frac{1}{4} (\frac{1}{2} m v^2)$.Since $K = \frac{1}{2} m v^2$,we have $K' = \frac{K}{4}$.

$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

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