The key idea is that, instead of reflecting the line $AY$ off of $BC$, we will reflect $ABC$ about $BC$ and extend $AY$ beyond $\triangle ABC$. We keep doing this until the extension of $AY$ hits a vertex of one of our reflected triangles. This is illustrated in the diagram below: We can calculate that the line $AY$ has slope $\frac{\frac{3 \sqrt{3}}{2}}{\frac{7}{2}}=\frac{3 \sqrt{3}}{7}$, so that (as indicated in the diagram), $AY$ first intersects a vertex at the point $\left(\frac{35}{2}, \frac{15 \sqrt{3}}{2}\right)^{2}$. To get there, it has to travel through 2 horizontal lines, 1 upward sloping line, and 4 downward sloping lines, so it bounces $2+1+4=7$ times total.