1. Initial kinetic energy

Let the particle be projected with speed $u$.

Then its initial kinetic energy is $K = \frac{1}{2}mu^2.$

2. Velocity components

The particle is projected at an angle $60^\circ$ to the horizontal, so: $u_x = u\cos 60^\circ = \frac{u}{2},$ $u_y = u\sin 60^\circ = \frac{\sqrt{3}u}{2}.$

3. 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 the speed at the highest point is $v = u_x = \frac{u}{2}.$

4. Kinetic energy at the highest point

Therefore, $K_{\text{top}} = \frac{1}{2}m\left(\frac{u}{2}\right)^2 = \frac{1}{2}m\cdot \frac{u^2}{4} = \frac{1}{4}\left(\frac{1}{2}mu^2\right).$

Since $\frac{1}{2}mu^2 = K$, $K_{\text{top}} = \frac{K}{4}.$

5. Option check

• A: $K/2$ ❌
• B: $K$ ❌
• C: Zero ❌
• D: $K/4$ ✅

Hence, the correct answer is D.