check the picture below, so, what is from the A coordinates, what's the y-coordinate then.
well, we can get that off either equation, if we just know A coordinates, so let's see what "x" is at those A and B points.
[tex]\bf \begin{cases}
y=-0.3x^2+3x\\ ----------\\
-2x+5.5y=19.5\\\\
y=\cfrac{19.5+2x}{5.5}
\end{cases}\implies -0.3x^2+3x=\cfrac{19.5+2x}{5.5}
\\\\\\
-1.65x^2+16.5x=19.5+2x\implies 0=1.65x^2-14.5x+19.5
\\\\\\
\textit{using the quadratic formula}\quad x=\cfrac{14.5\pm\sqrt{210.25-4(1.65)(19.5)}}{2(1.65)}
\\\\\\
x\approx
\begin{cases}
7.13\\
1.66
\end{cases}\textit{ so those are A and B, who is who?, dunno, let's check}[/tex]
[tex]\bf y=-0.3x^2+3x\qquad x=7.13\qquad y=-0.3(7.13)^2+3(7.13)
\\\\\\
y\approx 6.14\\\\
-------------------------------\\\\
y=-0.3x^2+3x\qquad x=1.66\qquad y=-0.3(1.66)^2+3(1.66)
\\\\\\
y\approx 4.15[/tex]
so, as you can see, clearly 6.14 is higher, and thus is B.
thus A is 4.15, so the y-coordinate for point A is 4.15, and that's the distance from the beam hitting the fountain at A, to the ground.
