Check the picture below, that'd be more or less the pattern of the sprinklinkg, so when we find it's vertex, we'd find how far along the ground at the maximum and its maximum height.
[tex]h(x)=160x-16x^2\implies h(x)=-16x^2+160x+0 \\\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(x)=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+160}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\left( -\cfrac{160}{2(-16)}~~,~~0-\cfrac{160^2}{4(-16)} \right)\implies \left( \cfrac{160}{32}~~,~~\cfrac{25600}{64} \right) \\\\\\ (\stackrel{\stackrel{\textit{along the ground}}{\downarrow }}{5}~~,~~\underset{\stackrel{\uparrow }{\textit{maximum height}}}{400})[/tex]
well, as you may notice from the picture, the water will hit the ground again when h(x) = 0 or namely y = 0, so let's do that to find out
[tex]h(x)= 160x-16x^2\implies \stackrel{h(x)}{0}= 160x-16x^2 \\\\\\ \stackrel{\textit{common factoring}}{0=16x(10-x)} \implies x = \begin{cases} 0&\leftarrow \textit{hits the ground first}\\ 10&\leftarrow \textit{hits \underline{the ground again}} \end{cases}[/tex]
