The TSR solution for that bit is overly perverse. All you need is a one-line observation that $\dot{\theta}^2 > 0$ so there can only be a sudden change in direction for $\theta$ to vary in a collision. Since the first collision happens at $\theta = 0$ then the second must be at $\theta = \pi$.

But to answer your question, the TSR solution says that just before the first collision ($\theta = 0$) the $y$-coordinate of $A$ from part (i) is $a\dot{\theta}\cos 0 = a \dot{\theta} = a \cdot - \frac{u}{a\sqrt{3}} = -\frac{u}{\sqrt{3}}$. So the *speed* of $A$ in the y-direction at $\theta = 0$ is $\frac{u}{\sqrt{3}}$.