5.1 m
Given:
The initial vertical velocity component is determined by: \[ v_{0y} = v_0 \sin \theta \] Using the provided values: \[ v_{0y} = 20 \, \text{m/s} \times \sin(30^\circ) = 20 \, \text{m/s} \times 0.5 = 10 \, \text{m/s} \]
The maximum height \( H \) is calculated using: \[ H = \frac{v_{0y}^2}{2g} \] Substituting the values: \[ H = \frac{(10 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{100}{19.6} = 5.1 \, \text{m} \]
The projectile achieves a maximum height of \( \boxed{5.1 \, \text{m}} \).