Let's break down the problem step by step:

The ball is projected with kinetic energy

$$\mathrm{KE} = \frac{1}{2} m v^2,$$

where $$v$$ is the initial speed.

The ball is launched at an angle of $$60^\circ$$. So, its initial velocity components are:

Horizontal: $$v_x = v \cos(60^\circ) = \frac{v}{2}$$

Vertical: $$v_y = v \sin(60^\circ)$$

At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., $$v_y = 0$$), but the horizontal component remains unchanged.

Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:

$$\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2$$

Substitute $$v_x = \frac{v}{2}$$:

$$\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$$

Notice that the initial kinetic energy is

$$\mathrm{KE} = \frac{1}{2} m v^2.$$

So we can write:

$$\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}$$

Thus, the kinetic energy at the highest point is $$\frac{\mathrm{KE}}{4}$$.

Answer: Option C.