1. Let the initial speed be $u$.

   The initial kinetic energy is

   $$\mathrm{KE}=\frac{1}{2}mu^2$$

2. Resolve the velocity into components.

   Since the ball is projected at $60^\circ$,

   $$u_x=u\cos 60^\circ=\frac{u}{2}, \qquad u_y=u\sin 60^\circ=\frac{\sqrt{3}u}{2}$$

3. Velocity at the highest point.

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

   So speed at the highest point is

   $$v= u_x = \frac{u}{2}$$

4. Kinetic energy at the highest point.

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

5. Compare with initial kinetic energy.

   Since

   $$\mathrm{KE}=\frac{1}{2}mu^2$$

   therefore,

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

6. Option check.

   • A: $\dfrac{\mathrm{KE}}{2}$ ❌
   • B: $\dfrac{\mathrm{KE}}{8}$ ❌
   • C: $\dfrac{\mathrm{KE}}{4}$ ✅
   • D: $\dfrac{\mathrm{KE}}{16}$ ❌

Hence, the correct answer is Option C.