1. Let the initial speed of projection be $u$.

2. Kinetic energy at the point of projection

At launch, the speed is $u$, so $K_1 = \frac{1}{2}mu^2$

3. Speed at the highest point

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

Given angle of projection $=30^\circ$, $u_x = u\cos 30^\circ = u\cdot \frac{\sqrt{3}}{2}$

So, speed at the highest point is $v_{top} = u\cos 30^\circ = \frac{\sqrt{3}}{2}u$

4. Kinetic energy at the highest point

$K_2 = \frac{1}{2}m\left(\frac{\sqrt{3}}{2}u\right)^2$ $K_2 = \frac{1}{2}m\cdot \frac{3u^2}{4}$ $K_2 = \frac{3}{8}mu^2$

5. Required ratio

$K_1 : K_2 = \frac{1}{2}mu^2 : \frac{3}{8}mu^2$

Cancel $mu^2$: $\frac{1}{2} : \frac{3}{8}$

Multiply both terms by $8$: $4 : 3$

6. Correct option

$\boxed{4:3}$ So, the correct answer is Option C.