A ball is projected with kinetic energy E, at an angle of $$60^\circ$$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become:

We have a ball projected with kinetic energy $$E$$ at an angle of $$60^\circ$$ to the horizontal. At the highest point of its flight, the vertical component of velocity becomes zero, and only the horizontal component remains.

The horizontal component of velocity is $$v_x = v\cos 60^\circ = \frac{v}{2}$$, where $$v$$ is the initial speed. The initial kinetic energy is $$E = \frac{1}{2}mv^2$$.

At the highest point, the kinetic energy is $$E' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2}m \cdot \frac{v^2}{4} = \frac{1}{4}\left(\frac{1}{2}mv^2\right) = \frac{E}{4}$$.

Hence, the correct answer is Option C.