1. **State the problem:** We want to find the maximum height of an object launched upward with height function $$s(t) = -16t^2 + 64t + 80$$ where $t$ is time in seconds. 2. **Formula and rules:** This is a quadratic function in the form $$s(t) = at^2 + bt + c$$ with $a = -16$, $b = 64$, and $c = 80$. Since $a < 0$, the parabola opens downward, so the vertex represents the maximum point. 3. **Find the time at maximum height:** The time $t$ at the vertex is given by $$t = -\frac{b}{2a} = -\frac{64}{2 \times (-16)} = -\frac{64}{-32} = 2$$ seconds. 4. **Calculate the maximum height:** Substitute $t=2$ into $s(t)$: $$s(2) = -16(2)^2 + 64(2) + 80 = -16(4) + 128 + 80 = -64 + 128 + 80$$ 5. **Simplify:** $$-64 + 128 + 80 = 64 + 80 = 144$$ feet. **Final answer:** The maximum height reached by the object is **144 feet**.