1. **State the problem:** We have the height equation of an object thrown upward: $$h = -16t^2 + 157t + 3$$ We need to find: - When the height $h$ is 280 feet. - When the object hits the ground (height $h=0$). 2. **Formula and rules:** This is a quadratic equation in $t$. To find $t$ for a given height $h$, solve: $$-16t^2 + 157t + 3 = h$$ Rearranged as: $$-16t^2 + 157t + (3 - h) = 0$$ Use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = -16$, $b = 157$, and $c = 3 - h$. 3. **Find when height is 280 feet:** Set $h=280$: $$-16t^2 + 157t + 3 - 280 = 0 \Rightarrow -16t^2 + 157t - 277 = 0$$ Apply quadratic formula: $$t = \frac{-157 \pm \sqrt{157^2 - 4(-16)(-277)}}{2(-16)}$$ Calculate discriminant: $$157^2 = 24649$$ $$4 \times 16 \times 277 = 17728$$ Discriminant: $$24649 - 17728 = 6921$$ Square root: $$\sqrt{6921} \approx 83.22$$ Calculate $t$ values: $$t = \frac{-157 \pm 83.22}{-32}$$ First root: $$t = \frac{-157 + 83.22}{-32} = \frac{-73.78}{-32} = 2.31$$ Second root: $$t = \frac{-157 - 83.22}{-32} = \frac{-240.22}{-32} = 7.51$$ So, the object reaches 280 feet at approximately $t=2.31$ seconds and $t=7.51$ seconds. 4. **Find when it hits the ground:** Set $h=0$: $$-16t^2 + 157t + 3 = 0$$ Apply quadratic formula with $a=-16$, $b=157$, $c=3$: $$t = \frac{-157 \pm \sqrt{157^2 - 4(-16)(3)}}{2(-16)}$$ Calculate discriminant: $$157^2 = 24649$$ $$4 \times 16 \times 3 = 192$$ Discriminant: $$24649 + 192 = 24841$$ Square root: $$\sqrt{24841} \approx 157.58$$ Calculate $t$ values: $$t = \frac{-157 \pm 157.58}{-32}$$ First root: $$t = \frac{-157 + 157.58}{-32} = \frac{0.58}{-32} = -0.018$$ (discard negative time) Second root: $$t = \frac{-157 - 157.58}{-32} = \frac{-314.58}{-32} = 9.83$$ So, the object hits the ground at approximately $t=9.83$ seconds. **Final answers:** - Height 280 feet at $t \approx 2.31$ seconds and $t \approx 7.51$ seconds. - Hits ground at $t \approx 9.83$ seconds.