A particle having kinetic energy K is project | vedclass

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(C) Let the initial velocity of projection be $u$. The initial kinetic energy is given by $KE_1 = \frac{1}{2} m u^2 = K$.At the highest point of the t

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$K / 4$

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(C) Let the initial velocity of projection be $u$. The initial kinetic energy is given by $KE_1 = \frac{1}{2} m u^2 = K$.At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains $u \cos 	heta$.Therefore,the kinetic energy at the highest point $KE_2$ is:$KE_2 = \frac{1}{2} m (u \cos 	heta)^2$$KE_2 = (\frac{1}{2} m u^2) \cos^2 	heta$Substituting $KE_1 = K$ and $	heta = 60^{\circ}$:$KE_2 = K \cos^2 60^{\circ}$Since $\cos 60^{\circ} = \frac{1}{2}$,we have:$KE_2 = K (\frac{1}{2})^2 = \frac{K}{4}$

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

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