1. The problem is to analyze the function $h(t) = -4.9t^2 + 30t + 55$, which models the height of an object over time $t$. 2. This is a quadratic function in the form $h(t) = at^2 + bt + c$ where $a = -4.9$, $b = 30$, and $c = 55$. 3. Since $a < 0$, the parabola opens downward, meaning the object reaches a maximum height at the vertex. 4. The vertex $t$-coordinate is given by the formula $$t = -\frac{b}{2a} = -\frac{30}{2 \times -4.9} = \frac{30}{9.8} = 3.06.$$ 5. Substitute $t=3.06$ back into $h(t)$ to find the maximum height: $$h(3.06) = -4.9(3.06)^2 + 30(3.06) + 55.$$ Calculate step-by-step: $$3.06^2 = 9.36,$$ $$-4.9 \times 9.36 = -45.86,$$ $$30 \times 3.06 = 91.8,$$ So, $$h(3.06) = -45.86 + 91.8 + 55 = 100.94.$$ 6. The maximum height is approximately $100.94$ units at $t = 3.06$ seconds. 7. The initial height at $t=0$ is $h(0) = 55$. 8. The object will eventually hit the ground when $h(t) = 0$. Solve: $$-4.9t^2 + 30t + 55 = 0.$$ 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 discriminant: $$30^2 = 900,$$ $$4 \times -4.9 \times 55 = -1078,$$ $$900 - (-1078) = 1978.$$ Square root: $$\sqrt{1978} \approx 44.47.$$ So, $$t = \frac{-30 \pm 44.47}{-9.8}.$$ Two solutions: $$t_1 = \frac{-30 + 44.47}{-9.8} = \frac{14.47}{-9.8} = -1.48$$ (discard negative time), $$t_2 = \frac{-30 - 44.47}{-9.8} = \frac{-74.47}{-9.8} = 7.6.$$ 9. The object hits the ground at approximately $t = 7.6$ seconds. Summary: - The graph is a downward-opening parabola. - Maximum height is about $100.94$ at $t=3.06$. - Initial height is $55$. - Hits ground at $t=7.6$ seconds.