Where x is the horizontal distance in feet from the point at which the ball is thrown.How far from the child does the ball strike the ground?

To calculate the maximum distance that the ball takes, we can use the first and second derivatives of y to find out this. We have:
[tex]\begin{gathered} y=-\frac{1}{14}x^2+4x+3 \\ \Rightarrow y^{\prime}=-\frac{2}{14}x+4=-\frac{1}{7}x+4 \\ \Rightarrow y^{\doubleprime}=-\frac{1}{7} \end{gathered}[/tex]Using the second derivative criterion, we have that y'' < 0, therefore, we have a maximum in the root of the first derivative. For that, we get the following:
[tex]\begin{gathered} y^{\prime}=0 \\ \Rightarrow-\frac{1}{7}x+4=0 \\ \Rightarrow\frac{1}{7}x=4 \\ \Rightarrow x=7\cdot4=28 \\ x=28 \end{gathered}[/tex]Therefore, at x=28 is where the maximum distance is. Now we only substitute x=28 in y to find out:
[tex]\begin{gathered} y=-\frac{1}{14}(28)^2+4(28)+3 \\ \Rightarrow y=-\frac{1}{14}(784)+112+3 \\ \Rightarrow y=-56+112+3=59 \\ y=59 \end{gathered}[/tex]Finally, we have that the ball will go 59 feet from the child