1. **State the problem:** We have the height function of a ball thrown upwards given by $$h(t) = 2 + 20t - 4.9t^2$$ where $h(t)$ is the height in meters and $t$ is the time in seconds. 2. **Part (a): Find the height at $t=3$ seconds.** Substitute $t=3$ into the function: $$h(3) = 2 + 20(3) - 4.9(3)^2 = 2 + 60 - 4.9 \times 9 = 2 + 60 - 44.1 = 17.9$$ So, the height at 3 seconds is $17.9$ meters. 3. **Part (b): Find the times when the height is 6 meters.** Set $h(t) = 6$: $$6 = 2 + 20t - 4.9t^2$$ Rearranged: $$0 = 2 + 20t - 4.9t^2 - 6 = -4.9t^2 + 20t - 4$$ Multiply both sides by $-1$ for easier handling: $$4.9t^2 - 20t + 4 = 0$$ Use the quadratic formula: $$t = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 4.9 \times 4}}{2 \times 4.9} = \frac{20 \pm \sqrt{400 - 78.4}}{9.8} = \frac{20 \pm \sqrt{321.6}}{9.8}$$ Calculate the square root: $$\sqrt{321.6} \approx 17.93$$ So, $$t_1 = \frac{20 - 17.93}{9.8} = \frac{2.07}{9.8} \approx 0.211$$ $$t_2 = \frac{20 + 17.93}{9.8} = \frac{37.93}{9.8} \approx 3.87$$ The ball is at 6 meters at approximately $0.21$ seconds and $3.87$ seconds. 4. **Part (c): Find the maximum height.** Since the function is a downward opening parabola, the maximum height is at the vertex. The vertex time is: $$t = -\frac{b}{2a} = -\frac{20}{2 \times (-4.9)} = \frac{20}{9.8} \approx 2.04$$ Calculate the height at $t=2.04$: $$h(2.04) = 2 + 20(2.04) - 4.9(2.04)^2 = 2 + 40.8 - 4.9 \times 4.16 = 2 + 40.8 - 20.38 = 22.42$$ The maximum height is approximately $22.42$ meters at $2.04$ seconds.