To determine the maximum height reached by a projectile, we can use the following formula:
\[ H = \frac{v_y^2}{2g} \]
where:
- \( H \) is the maximum height,
- \( v_y \) is the initial vertical velocity,
- \( g \) is the acceleration due to gravity (approx. \( 9.81 , \text{m/s}^2 \)).
First, we need to find the initial vertical velocity \( v_y \). The initial velocity \( v_0 \) is given as 20 m/s, and we can find \( v_y \) using the angle of projection \( \theta = 60^\circ \):
\[ v_y = v_0 \sin(\theta) \]
Calculating \( v_y \):
\[ v_y = 20 \sin(60^\circ) \]
We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Therefore:
\[ v_y = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} , \text{m/s} \]
Now we can substitute \( v_y \) into the height formula:
\[ H = \frac{(10\sqrt{3})^2}{2 \cdot 9.81} \]
Calculating \( (10\sqrt{3})^2 \):
\[ (10\sqrt{3})^2 = 100 \cdot 3 = 300 \]
Now substituting this into the height formula:
\[ H = \frac{300}{2 \cdot 9.81} = \frac{300}{19.62} \approx 15.29 \text{ meters} \]
Thus, the maximum height reached by the projectile is approximately \( \boxed{15.29} \) meters.