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

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$E / 4$

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(C) At the ground,the initial kinetic energy is given by $E = \frac{1}{2} m u^{2}$,where $u$ 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 only the horizontal component,which is $v_x = u \cos 	heta$.Given the angle of projection $	heta = 60^{\circ}$,the horizontal velocity at the highest point is $v_x = u \cos 60^{\circ} = u 	imes \frac{1}{2} = \frac{u}{2}$.The kinetic energy at the highest point $E^{\prime}$ is given by $E^{\prime} = \frac{1}{2} m v_x^{2}$.Substituting the value of $v_x$,we get $E^{\prime} = \frac{1}{2} m \left( \frac{u}{2} ight)^{2} = \frac{1}{2} m \left( \frac{u^{2}}{4} ight) = \frac{1}{4} \left( \frac{1}{2} m u^{2} ight)$.Since $E = \frac{1}{2} m u^{2}$,we have $E^{\prime} = \frac{E}{4}$.

$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

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