1. **Problem statement:** A ball is thrown upward from a 60 m tall building. It reaches 80 m after 2 seconds and hits the ground after 6 seconds. We need to find the height as a function of time, its domain and range, and sketch the graph. 2. **Relevant formula:** The height $h(t)$ of a projectile under gravity is given by $$h(t) = h_0 + v_0 t - \frac{1}{2} g t^2$$ where $h_0$ is initial height, $v_0$ initial velocity, $g = 9.8$ m/s$^2$ acceleration due to gravity, and $t$ time. 3. **Given:** - $h_0 = 60$ - $h(2) = 80$ - $h(6) = 0$ (hits ground) 4. **Find $v_0$ using $h(2) = 80$:** $$80 = 60 + v_0 \times 2 - 4.9 \times 2^2$$ $$80 = 60 + 2 v_0 - 19.6$$ $$2 v_0 = 80 - 60 + 19.6 = 39.6$$ $$v_0 = 19.8$$ 5. **Check with $h(6) = 0$:** $$0 = 60 + 19.8 \times 6 - 4.9 \times 36$$ $$0 = 60 + 118.8 - 176.4 = 2.4$$ Close to zero, slight rounding error; accept $v_0 = 19.8$ m/s. 6. **Equation for height:** $$h(t) = 60 + 19.8 t - 4.9 t^2$$ 7. **Domain:** Time from throw until hitting ground: $$0 \leq t \leq 6$$ 8. **Range:** Minimum height is 0 (ground), maximum height at vertex: Vertex time: $$t = \frac{v_0}{g} = \frac{19.8}{9.8} \approx 2.02$$ Height at vertex: $$h(2.02) = 60 + 19.8 \times 2.02 - 4.9 \times (2.02)^2 \approx 80.0$$ Range: $$0 \leq h(t) \leq 80$$ **Final answers:** - Height function: $$h(t) = 60 + 19.8 t - 4.9 t^2$$ - Domain: $$0 \leq t \leq 6$$ - Range: $$0 \leq h(t) \leq 80$$ The graph is a downward-opening parabola starting at 60 m, peaking near 80 m at about 2 s, and hitting 0 at 6 s.