A particle is projected at 60^circ to the hor | vedclass

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(D) Let $u$ be the initial velocity with which the particle is thrown and $m$ be the mass of the particle. The initial kinetic energy is given by $K =

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$\frac{K}{4}$

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(D) Let $u$ be the initial velocity with which the particle is thrown and $m$ be the mass of the particle. The initial kinetic energy is given by $K = \frac{1}{2}mu^2$.At the highest point of the trajectory, the vertical component of velocity becomes zero, and only the horizontal component of velocity remains.The horizontal component of velocity is $v_x = u \cos(60^\circ)$.Therefore, the kinetic energy at the highest point $K'$ is:$K' = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(u \cos(60^\circ))^2$$K' = \frac{1}{2}mu^2 \cos^2(60^\circ)$Since $\cos(60^\circ) = \frac{1}{2}$, we have $\cos^2(60^\circ) = \frac{1}{4}$.Substituting $K = \frac{1}{2}mu^2$, we get:$K' = K 	imes \frac{1}{4} = \frac{K}{4}$.

$A$ 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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