1. **State the problem:** Find the maximum height of the rocket given by the height function $$h(t) = -16t^2 + 96t + 112$$ where $t$ is time in seconds. 2. **Formula and rules:** The height function is a quadratic in the form $$h(t) = at^2 + bt + c$$ with $a = -16$, $b = 96$, and $c = 112$. Since $a < 0$, the parabola opens downward, so the vertex represents the maximum point. The time at which the maximum height occurs is given by the vertex formula: $$t = -\frac{b}{2a}$$ 3. **Calculate the time of maximum height:** $$t = -\frac{96}{2 \times -16} = -\frac{96}{-32} = 3$$ seconds. 4. **Calculate the maximum height by substituting $t=3$ into $h(t)$:** $$h(3) = -16(3)^2 + 96(3) + 112$$ $$= -16 \times 9 + 288 + 112$$ $$= -144 + 288 + 112$$ $$= 256$$ feet. 5. **Interpretation:** The rocket reaches its maximum height of 256 feet at 3 seconds after launch. Final answer: The maximum height is **256 feet** at **3 seconds**.