1. **State the problem:** We have a quadratic function modeling the height of a tennis ball over time: $$f(t) = -16t^2 + 46t + 6$$ where $f(t)$ is the height in feet and $t$ is the time in seconds. We need to find: - How long it takes for the ball to hit the ground (i.e., when height is 0). - When the ball is 2 feet from the ground (i.e., when height is 2). 2. **Formula and rules:** To find when the ball hits the ground, solve for $t$ when $f(t) = 0$. To find when the ball is 2 feet high, solve for $t$ when $f(t) = 2$. The quadratic formula for $ax^2 + bx + c = 0$ is: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Find when the ball hits the ground:** Set $f(t) = 0$: $$-16t^2 + 46t + 6 = 0$$ Here, $a = -16$, $b = 46$, $c = 6$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 46^2 - 4(-16)(6) = 2116 + 384 = 2500$$ Calculate $t$: $$t = \frac{-46 \pm \sqrt{2500}}{2(-16)} = \frac{-46 \pm 50}{-32}$$ Two solutions: $$t_1 = \frac{-46 + 50}{-32} = \frac{4}{-32} = -\frac{1}{8}$$ $$t_2 = \frac{-46 - 50}{-32} = \frac{-96}{-32} = 3$$ Since time cannot be negative, the ball hits the ground at: $$t = 3 \text{ seconds}$$ 4. **Find when the ball is 2 feet from the ground:** Set $f(t) = 2$: $$-16t^2 + 46t + 6 = 2$$ Simplify: $$-16t^2 + 46t + 4 = 0$$ Calculate the discriminant: $$\Delta = 46^2 - 4(-16)(4) = 2116 + 256 = 2372$$ Calculate $t$: $$t = \frac{-46 \pm \sqrt{2372}}{2(-16)}$$ Approximate $\sqrt{2372} \approx 48.7$: $$t_1 = \frac{-46 + 48.7}{-32} = \frac{2.7}{-32} = -0.084$$ $$t_2 = \frac{-46 - 48.7}{-32} = \frac{-94.7}{-32} = 2.96$$ Ignoring negative time, the ball is 2 feet high at approximately: $$t = 2.96 \text{ seconds}$$ **Final answers:** - The ball hits the ground after $3$ seconds. - The ball is 2 feet from the ground at approximately $2.96$ seconds.