1. **Problem Statement:** Given the height function of a bungee jumper $$h(t) = -0.5t^2 + v_0t + h_0$$ with $$v_0=0$$ and $$h_0=210$$ meters, analyze the domain, range, vertex, maximum height, times for specific heights, and when the jumper touches the river. 2. **Formula and Rules:** The function is quadratic, $$h(t) = -0.5t^2 + 0 imes t + 210 = -0.5t^2 + 210$$. - Domain: Time $$t \\geq 0$$ since time cannot be negative. - Range: Values of $$h(t)$$ from the maximum height down to the river surface. - Vertex formula: $$t = -\frac{b}{2a}$$ for $$ax^2 + bx + c$$. 3. **Domain and Range:** - Domain: $$[0, t_{river}]$$ where $$t_{river}$$ is when $$h(t) = 0$$. - Range: $$[0, 210]$$ meters physically, but since the jumper falls, the minimum height is 0 (river surface). - Physical significance: Domain represents the time interval of the jump; range represents possible heights. 4. **Vertex:** - Here, $$a = -0.5$$, $$b=0$$, $$c=210$$. - Vertex time: $$t = -\frac{0}{2(-0.5)} = 0$$ seconds. - Vertex height: $$h(0) = 210$$ meters. - Interpretation: The jumper starts at maximum height 210 m at $$t=0$$. 5. **Maximum Height and Time:** - Maximum height is at vertex: $$210$$ meters at $$t=0$$ seconds. - Since $$a<0$$, parabola opens downward, so height decreases over time. 6. **Time when height is 11 m:** Solve $$-0.5t^2 + 210 = 11$$ $$-0.5t^2 = 11 - 210 = -199$$ $$t^2 = \frac{199}{0.5} = 398$$ $$t = \sqrt{398} \approx 19.95$$ seconds. 7. **Height after 20 seconds:** $$h(20) = -0.5(20)^2 + 210 = -0.5(400) + 210 = -200 + 210 = 10$$ meters. - This means after 20 seconds, the jumper is 10 meters above the river. 8. **Time when jumper touches river:** Set $$h(t) = 0$$: $$-0.5t^2 + 210 = 0$$ $$-0.5t^2 = -210$$ $$t^2 = 420$$ $$t = \sqrt{420} \approx 20.49$$ seconds. **Final answers:** - Domain: $$[0, 20.49]$$ seconds - Range: $$[0, 210]$$ meters - Vertex: $$(0, 210)$$ - Maximum height: 210 m at $$t=0$$ - Height 11 m at $$t \approx 19.95$$ seconds - Height after 20 s: 10 m - Touches river at $$t \approx 20.49$$ seconds