Solution:

At the highest point of its trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged.

Initial kinetic energy \( K \) is given by:

\[
K = \frac{1}{2} m u^2
\]

At the highest point, only the horizontal component of the velocity \( u \cos 60^\circ \) remains. Therefore, the kinetic energy at the highest point is due to this horizontal velocity:

\[
K_{\text{highest}} = \frac{1}{2} m (u \cos 60^\circ)^2
\]

Since \( \cos 60^\circ = \frac{1}{2} \):

\[
K_{\text{highest}} = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2} m u^2 = \frac{K}{2}
\]

Thus, the kinetic energy at the highest point is \( \frac{K}{2} \).