1. **State the problem:** We need to find how long the catcher has to get in position to catch the ball before it hits the ground. The height of the ball is given by the quadratic function $$b(t) = 80t - 16t^2 + 3.5$$ where $t$ is time in seconds. 2. **Set the height to zero to find when the ball hits the ground:** $$80t - 16t^2 + 3.5 = 0$$ 3. **Rewrite the equation in standard quadratic form:** $$-16t^2 + 80t + 3.5 = 0$$ 4. **Identify coefficients:** $$a = -16, \quad b = 80, \quad c = 3.5$$ 5. **Use the quadratic formula:** $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 6. **Calculate the discriminant:** $$b^2 - 4ac = 80^2 - 4(-16)(3.5) = 6400 + 224 = 6624$$ 7. **Substitute values into the quadratic formula:** $$t = \frac{-80 \pm \sqrt{6624}}{2(-16)}$$ 8. **Simplify the denominator:** $$2(-16) = -32$$ 9. **Calculate the square root:** $$\sqrt{6624} \approx 81.39$$ 10. **Find the two possible times:** $$t_1 = \frac{-80 + 81.39}{-32} = \frac{1.39}{-32} \approx -0.043 \quad (\text{not physically meaningful since time cannot be negative})$$ $$t_2 = \frac{-80 - 81.39}{-32} = \frac{-161.39}{-32} \approx 5.04$$ 11. **Interpret the result:** The ball hits the ground after approximately $5.04$ seconds. 12. **Round to the nearest second:** The catcher has about **5 seconds** to get in position to catch the ball before it hits the ground.