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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 traj

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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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