The maximum height of a projectile is determined by its vertical motion. At its peak, the projectile's vertical velocity is zero. The equation governing this is:
\[v_y^2 = u_y^2 - 2gH_{\text{max}},\]
where \( u_y = u \sin \theta \) represents the initial vertical velocity, \( v_y = 0 \) at maximum height, \( g \) is the acceleration due to gravity, and \( H_{\text{max}} \) is the maximum height.
Rearranging to solve for \( H_{\text{max}} \):
\[H_{\text{max}} = \frac{u_y^2}{2g} = \frac{(u \sin \theta)^2}{2g}.\]
Consider two projectiles with projection angles:
\[\theta_1 = 30^\circ \quad \text{and} \quad \theta_2 = 60^\circ.\]
For the first projectile:
\[H_1 = \frac{(u \sin 30^\circ)^2}{2g}.\]
For the second projectile:
\[H_2 = \frac{(u \sin 60^\circ)^2}{2g}.\]
The ratio of their maximum heights is:
\[\frac{H_1}{H_2} = \frac{(u \sin 30^\circ)^2}{(u \sin 60^\circ)^2}.\]
Canceling the common term \( u^2 \) and substituting \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[\frac{H_1}{H_2} = \frac{\left(\frac{1}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2}.\]
Simplifying the expression:
\[\frac{H_1}{H_2} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.\]
Therefore, the ratio of the maximum heights is:
\[H_1 : H_2 = 1 : 3.\]
Final Answer:
\[\boxed{1 : 3}\]