Given that the ball travels 10 ft horizontally, then the maximum is reached at half of this distance, that is 5 ft. This means that the parabola has a maximum at (5, 25).
The equation of a parabola in vertex form is:
![y=a(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/college/97p0xsjs0cwme4ddvwkim2cbbqprhnlhsv.png)
where (h, k) is the vertex.
Substituting with the vertex (5, 25) and the point (0,0) we get:
![\begin{gathered} 0=a(0-5)^2+25 \\ 0=a\cdot25+25 \\ -25=a\cdot25 \\ -(25)/(25)=a \\ -1=a \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wkzffsgngz0h95kc86g6ore25gn90cq7s3.png)
And the function that models the path traveled by the ball is:
![y=-(x-5)^2+25](https://img.qammunity.org/2023/formulas/mathematics/college/swh6fp2a155q3uj037dw6c2cl0yeb8i15w.png)
where y is the vertical distance, or height, in ft, and x is the horizontal distance, also in ft.