1. **State the problem:** A projectile is launched at an angle of 40 degrees with an initial speed of 20 m/s. We need to find its maximum height. 2. **Formula used:** The maximum height $H$ of a projectile launched at an angle $\theta$ with initial speed $v_0$ is given by: $$H = \frac{v_0^2 \sin^2(\theta)}{2g}$$ where $g = 9.8$ m/s$^2$ is the acceleration due to gravity. 3. **Calculate the vertical component of velocity:** $$v_{0y} = v_0 \sin(\theta) = 20 \times \sin(40^\circ)$$ Using $\sin(40^\circ) \approx 0.6428$: $$v_{0y} = 20 \times 0.6428 = 12.856 \text{ m/s}$$ 4. **Calculate maximum height:** $$H = \frac{(12.856)^2}{2 \times 9.8} = \frac{165.3}{19.6} = 8.44 \text{ meters}$$ 5. **Answer:** The maximum height reached by the projectile is approximately **8.44 meters**.