A body of mass m is projected from the ground with an initial...

Let the initial speed of the body be $u$. The initial kinetic energy is given by:
$K = \dfrac{1}{2}mu^2$

The angle of projection with the vertical is $30^{\circ}$. Therefore, the angle of projection with the horizontal is $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

At the highest point of the trajectory, the vertical component of velocity is zero. The speed of the body is equal to the horizontal component of its initial velocity, which remains constant.

Speed at highest point, $v = u \cos 60^{\circ} = u \times \dfrac{1}{2} = \dfrac{u}{2}$

The kinetic energy at the highest point is:
$K' = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dfrac{u}{2}\right)^2 = \dfrac{1}{4}\left(\dfrac{1}{2}mu^2\right)$

$\Rightarrow K' = \dfrac{K}{4}$

Answer: $\dfrac{K}{4}$