Bungee Height Ceae75
1. **Problem Statement:** We analyze the quadratic height function of a bungee jumper given by $$h(t) = -0.5t^2 + v_0 t + h_0$$ with $$v_0 = 0$$ and $$h_0 = 210$$ meters.
2. **Domain and Range:**
- The domain of $$h(t)$$ is the set of all possible times $$t$$ starting from 0 (jump time) until the jumper reaches the river (height 0). So, $$\text{Domain} = [0, t_{river}]$$.
- The range is the set of heights from the minimum height (river level, 0) up to the maximum height reached.
- Physically, the domain represents the time interval during which the jumper is in the air, and the range represents the possible heights above the river surface.
3. **Vertex of the Parabola:**
- The vertex formula for $$h(t) = at^2 + bt + c$$ is $$t = -\frac{b}{2a}$$.
- Here, $$a = -0.5$$, $$b = 0$$, so $$t = -\frac{0}{2(-0.5)} = 0$$.
- The vertex is at $$t=0$$, $$h(0) = 210$$.
- The vertex represents the initial height and the maximum height since the parabola opens downward.
4. **Maximum Height and Time:**
- Since $$v_0=0$$, the jumper starts at maximum height $$210$$ meters at $$t=0$$.
- The maximum height is $$210$$ meters at $$t=0$$ seconds.
5. **Time to Reach Height 11m:**
- Solve $$h(t) = 11$$:
$$-0.5t^2 + 210 = 11$$
$$-0.5t^2 = 11 - 210 = -199$$
$$t^2 = \frac{199}{0.5} = 398$$
$$t = \sqrt{398} \approx 19.95$$ seconds.
- So, the jumper reaches 11 meters at approximately $$19.95$$ seconds.
6. **Height at 20 seconds:**
- Calculate $$h(20) = -0.5(20)^2 + 210 = -0.5(400) + 210 = -200 + 210 = 10$$ meters.
- This means after 20 seconds, the jumper is 10 meters above the river.
7. **Time to Touch the River:**
- Solve $$h(t) = 0$$:
$$-0.5t^2 + 210 = 0$$
$$-0.5t^2 = -210$$
$$t^2 = \frac{210}{0.5} = 420$$
$$t = \sqrt{420} \approx 20.49$$ seconds.
- The jumper touches the river at approximately $$20.49$$ seconds.
**Summary:**
- Domain: $$[0, 20.49]$$ seconds.
- Range: $$[0, 210]$$ meters.
- Vertex: $$(0, 210)$$ maximum height.
- Height 11m at $$t \approx 19.95$$ seconds.
- Height at 20s is 10m.
- Touches river at $$t \approx 20.49$$ seconds.