1. **State the problem:** We are given the height function of a watermelon launched upward as $$h = -16t^2 + 64t + 80$$ where $h$ is the height in feet and $t$ is the time in seconds. 2. **Identify the type of function:** This is a quadratic function representing the height of the watermelon over time. The parabola opens downward because the coefficient of $t^2$ is negative ($-16$). 3. **Find the vertex (maximum height):** The vertex of a parabola given by $$h = at^2 + bt + c$$ is at $$t = -\frac{b}{2a}$$. Here, $a = -16$, $b = 64$, so $$t = -\frac{64}{2 \times -16} = -\frac{64}{-32} = 2$$ seconds. 4. **Calculate the maximum height:** Substitute $t=2$ into the height equation: $$h = -16(2)^2 + 64(2) + 80 = -16 \times 4 + 128 + 80 = -64 + 128 + 80 = 144$$ feet. 5. **Find when the watermelon hits the ground:** Set $h=0$ and solve for $t$: $$0 = -16t^2 + 64t + 80$$ Divide both sides by $-16$: $$0 = \cancel{-16}t^2 \div \cancel{-16} - \frac{64}{-16}t - \frac{80}{-16}$$ $$0 = t^2 - 4t - 5$$ 6. **Factor the quadratic:** $$t^2 - 4t - 5 = (t - 5)(t + 1) = 0$$ 7. **Solve for $t$:** $$t - 5 = 0 \Rightarrow t = 5$$ $$t + 1 = 0 \Rightarrow t = -1$$ (discard negative time) So, the watermelon hits the ground at $t=5$ seconds. **Final answers:** - Maximum height is $$144$$ feet at $$t=2$$ seconds. - The watermelon hits the ground at $$t=5$$ seconds.