A body is projected at an angle of 60^{circ} | vedclass

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(D) Let the initial velocity of the body be $u$ and its mass be $m$. The initial kinetic energy is given by $X = \frac{1}{2}mu^2$.At the highest point

Vedclass · Accepted Answer

$\frac{X}{4}$

Vedclass · Answer

(D) Let the initial velocity of the body be $u$ and its mass be $m$. The initial kinetic energy is given by $X = \frac{1}{2}mu^2$.At the highest point of the trajectory, the vertical component of velocity becomes zero, and the horizontal component remains $u_x = u \cos 	heta$.Given the angle of projection $	heta = 60^{\circ}$, the horizontal velocity at the highest point is $u_x = u \cos 60^{\circ} = u 	imes \frac{1}{2} = \frac{u}{2}$.The kinetic energy at the highest point $K_h$ is given by $K_h = \frac{1}{2}m(u_x)^2$.Substituting the value of $u_x$, we get $K_h = \frac{1}{2}m(\frac{u}{2})^2 = \frac{1}{2}m(\frac{u^2}{4}) = \frac{1}{4}(\frac{1}{2}mu^2)$.Since $X = \frac{1}{2}mu^2$, we have $K_h = \frac{X}{4}$.

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal. If the initial kinetic energy of the body is $X$, then its kinetic energy at the highest point is

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