1. **State the problem:** We are given the height function of an object launched vertically as $$h(t) = -4.9t^2 + 30t + 55$$ where $t$ is time in seconds and $h(t)$ is height in meters. 2. **Identify the type of function:** This is a quadratic function of the form $$h(t) = at^2 + bt + c$$ with $a = -4.9$, $b = 30$, and $c = 55$. 3. **Find the vertex (maximum height):** Since $a < 0$, the parabola opens downward and the vertex represents the maximum height. The time at which maximum height occurs is given by $$t = -\frac{b}{2a} = -\frac{30}{2 \times -4.9} = \frac{30}{9.8} = 3.06 \text{ seconds (approx)}.$$ 4. **Calculate the maximum height:** Substitute $t = 3.06$ into $h(t)$: $$h(3.06) = -4.9(3.06)^2 + 30(3.06) + 55$$ $$= -4.9 \times 9.36 + 91.8 + 55$$ $$= -45.86 + 91.8 + 55 = 100.94 \text{ meters (approx)}.$$ 5. **Find the time when the object hits the ground:** Set $h(t) = 0$ and solve for $t$: $$-4.9t^2 + 30t + 55 = 0$$ Use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-30 \pm \sqrt{30^2 - 4(-4.9)(55)}}{2(-4.9)}$$ Calculate the discriminant: $$30^2 - 4(-4.9)(55) = 900 + 1078 = 1978$$ So, $$t = \frac{-30 \pm \sqrt{1978}}{-9.8}$$ Calculate $$\sqrt{1978} \approx 44.47$$ Two solutions: $$t_1 = \frac{-30 + 44.47}{-9.8} = \frac{14.47}{-9.8} = -1.48 \text{ (discard negative time)}$$ $$t_2 = \frac{-30 - 44.47}{-9.8} = \frac{-74.47}{-9.8} = 7.60 \text{ seconds (approx)}$$ 6. **Interpretation:** The object reaches maximum height of approximately 100.94 meters at 3.06 seconds and hits the ground at approximately 7.60 seconds.