the ball is modeled by the function \[ y=-\frac{32}{(20)^{2}} x^{2}+x+5 \] where $x$ is the distance in feet that the ball has traveled horizontally. (a) Find the maximum height attained by the ball. (Round your answer to three decimal places.) $\mathrm{ft}$ (b) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.) $\mathrm{ft}$

Step 1 ：Given the function \(y = -0.08x^2 + x + 5\), which models the path of the ball, where \(x\) is the distance in feet that the ball has traveled horizontally.

Step 2 ：For part (a), we need to find the maximum value of the function. This occurs at the vertex of the parabola. The x-coordinate of the vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is \(-b/2a\). Substituting the values from our equation, we find that the x-coordinate of the vertex is 6.25.

Step 3 ：We can substitute this value back into the equation to find the y-coordinate, which will be the maximum height. Doing so, we find that the maximum height is 8.125 feet.

Step 4 ：For part (b), we need to find when the ball hits the ground, i.e., when \(y = 0\). This requires solving the quadratic equation \(0 = -0.08x^2 + x + 5\) for \(x\).

Step 5 ：Solving this equation, we find that the ball hits the ground when it has traveled approximately -3.83 feet or 16.33 feet horizontally. However, since distance cannot be negative, we discard the negative root.

Step 6 ：Thus, the ball has traveled approximately 16.33 feet when it hits the ground.

Step 7 ：Final Answer: The maximum height attained by the ball is \(\boxed{8.125}\) feet. The horizontal distance the ball has traveled when it hits the ground is \(\boxed{16.3}\) feet.