1. **Problem Statement:** A football is punted into the air with height $h$ in meters after $t$ seconds given by the quadratic equation: $$h = -4.9t^2 + 24.5t$$ We will answer the sub-questions step-by-step. 2. **Formula and Important Rules:** - The height function is a quadratic in standard form $h = at^2 + bt + c$ where $a = -4.9$, $b = 24.5$, and $c = 0$. - The graph is a parabola opening downwards because $a < 0$. - The vertex gives the maximum height. - The roots (when $h=0$) give the times when the ball is on the ground. 3. **a) Sketch:** - The $t$-axis represents time in seconds. - The $h$-axis represents height in meters. - The parabola opens downward starting at $h=0$ when $t=0$. 4. **b) Height after 1 second:** $$h(1) = -4.9(1)^2 + 24.5(1) = -4.9 + 24.5 = 19.6$$ So, the ball is 19.6 meters high after 1 second. 5. **c) Initial height (at $t=0$):** $$h(0) = -4.9(0)^2 + 24.5(0) = 0$$ The ball was punted from ground level. 6. **d) When does the ball land?** Set $h=0$ and solve for $t$: $$-4.9t^2 + 24.5t = 0$$ Factor out $t$: $$t(-4.9t + 24.5) = 0$$ So, $t=0$ (launch time) or $$-4.9t + 24.5 = 0 \Rightarrow t = \frac{24.5}{4.9} = 5$$ The ball lands after 5 seconds. 7. **e) Time to reach maximum height:** The vertex time is given by: $$t = -\frac{b}{2a} = -\frac{24.5}{2 \times (-4.9)} = \frac{24.5}{9.8} = 2.5$$ It takes 2.5 seconds to reach maximum height. 8. **f) Maximum height:** Substitute $t=2.5$ into $h$: $$h(2.5) = -4.9(2.5)^2 + 24.5(2.5) = -4.9(6.25) + 61.25 = -30.625 + 61.25 = 30.625$$ Maximum height is 30.625 meters. 9. **g) Time ball is above 19.6 meters:** Set $h = 19.6$ and solve for $t$: $$-4.9t^2 + 24.5t = 19.6$$ Rearranged: $$-4.9t^2 + 24.5t - 19.6 = 0$$ Multiply both sides by -1 for easier calculation: $$4.9t^2 - 24.5t + 19.6 = 0$$ Use quadratic formula: $$t = \frac{24.5 \pm \sqrt{(-24.5)^2 - 4 \times 4.9 \times 19.6}}{2 \times 4.9}$$ Calculate discriminant: $$= \sqrt{600.25 - 384.16} = \sqrt{216.09} = 14.7$$ So, $$t_1 = \frac{24.5 - 14.7}{9.8} = \frac{9.8}{9.8} = 1$$ $$t_2 = \frac{24.5 + 14.7}{9.8} = \frac{39.2}{9.8} = 4$$ The ball is above 19.6 meters between $t=1$ and $t=4$ seconds. **Duration:** $4 - 1 = 3$ seconds. **Final answer:** The ball is above 19.6 meters for 3 seconds.