The maximum height $h_m$ reached by a projectile is given by the formula: $h_m = \frac{(v_0 \sin \theta_0)^2}{2g}$ where $v_0$ is the initial speed, $\theta_0$ is the angle of projection, and $g$ is the acceleration due to gravity .

Given: Initial speed, $v_0 = 20 \text{ m/s}$ Angle of projection, $\theta_0 = 30^{\circ}$

• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Substituting these values into the equation: $h_m = \frac{(20 \sin 30^{\circ})^2}{2 \times 10}$ $h_m = \frac{(20 \times 0.5)^2}{20}$ $h_m = \frac{(10)^2}{20}$ $h_m = \frac{100}{20} = 5 \text{ m}$

Thus, the maximum height attained by the ball is $5 \text{ m}$.