1. **Problem statement:** A water rocket is launched from the ground with an initial velocity of 32 m/s. It reaches a height of 44 m after 2 seconds and stays in the air for 6 seconds. We need to find the quadratic function modeling the height over time and state the domain and range. 2. **Formula used:** The height $h(t)$ of a projectile under gravity is modeled by a quadratic function: $$h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$$ where $g = 9.8$ m/s² (acceleration due to gravity), $v_0$ is the initial velocity, and $h_0$ is the initial height. 3. **Given values:** - Initial velocity $v_0 = 32$ m/s - Initial height $h_0 = 0$ m (launched from ground) - Height at $t=2$ s is 44 m - Total time in air is 6 s 4. **Write the general function:** $$h(t) = -4.9t^2 + 32t + 0 = -4.9t^2 + 32t$$ 5. **Check height at $t=2$ s:** $$h(2) = -4.9(2)^2 + 32(2) = -4.9 \times 4 + 64 = -19.6 + 64 = 44.4$$ This is approximately 44 m, confirming the model. 6. **Domain:** Time $t$ starts at 0 and ends when the rocket hits the ground again, which is at $t=6$ s. So, domain: $$0 \leq t \leq 6$$ 7. **Range:** The maximum height is the vertex of the parabola. The vertex time is: $$t_{vertex} = -\frac{b}{2a} = -\frac{32}{2 \times (-4.9)} = \frac{32}{9.8} \approx 3.27 \text{ s}$$ Calculate max height: $$h(3.27) = -4.9(3.27)^2 + 32(3.27) \approx -4.9(10.7) + 104.64 = -52.43 + 104.64 = 52.21 \text{ m}$$ So, range: $$0 \leq h(t) \leq 52.21$$ **Final answers:** - Quadratic function: $$h(t) = -4.9t^2 + 32t$$ - Domain: $$0 \leq t \leq 6$$ - Range: $$0 \leq h(t) \leq 52.21$$