Answer:
<h2>The second ball traveled more distance, horizontally and vertically.</h2>
Explanation:
The given function is
![f(x)=-0.032x(x-50)](https://tex.z-dn.net/?f=f%28x%29%3D-0.032x%28x-50%29)
To find the distances that the second football travels, we need to find the vertex of its movement, because it's movement has a parabola form. The quadratic expression is
![f(x)=-0.032x^{2} +1.6x](https://tex.z-dn.net/?f=f%28x%29%3D-0.032x%5E%7B2%7D%20%2B1.6x)
Where
and ![b=1.6](https://tex.z-dn.net/?f=b%3D1.6)
The vertex has coordinates of
, where
![h=-\frac{b}{2a}](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7Bb%7D%7B2a%7D)
Replacing values, we have
![h=-\frac{1.6}{2(-0.032)}=25](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7B1.6%7D%7B2%28-0.032%29%7D%3D25)
Then, ![k=f(h)](https://tex.z-dn.net/?f=k%3Df%28h%29)
![k=f(25)=-0.032(25)^{2} +1.6(25)\\k=-20+40=20](https://tex.z-dn.net/?f=k%3Df%2825%29%3D-0.032%2825%29%5E%7B2%7D%20%2B1.6%2825%29%5C%5Ck%3D-20%2B40%3D20)
Which means the maximum height of the second football is 20 yards. That means it travels 40 yards vertically.
Now, its horizontal distance can be found when ![f(x)=0](https://tex.z-dn.net/?f=f%28x%29%3D0)
![0=-0.032x^{2} +1.6x\\0=x(-0.032x+1.6)\\x_{1}=0\\ -0.032x_{2} +16=0\\x_{2}=\frac{-16}{-0.032}\\ x_{2}=500](https://tex.z-dn.net/?f=0%3D-0.032x%5E%7B2%7D%20%2B1.6x%5C%5C0%3Dx%28-0.032x%2B1.6%29%5C%5Cx_%7B1%7D%3D0%5C%5C%20-0.032x_%7B2%7D%20%2B16%3D0%5C%5Cx_%7B2%7D%3D%5Cfrac%7B-16%7D%7B-0.032%7D%5C%5C%20x_%7B2%7D%3D500)
So, its horizontal distance is 500 yards.
Comparing the distances between the footballs.
<h3>Ball 1</h3>
Horizontal distance of 30 yards.
Vertical distance of 30 yards.
<h3>Ball 2</h3>
Horizontal distance of 500 yards.
Vertical distance of 40 yards.
If we find their difference, it would be
Horizontal: 500 - 30 = 470 yards.
Vertical: 40 - 30 = 10 yards.
Therefore, the second ball traveled more distance, horizontally and vertically.