2. Given $a \geqslant b>0$, and $\sin \alpha$ is a root of the quadratic equation $a x^{2}+b x-b=0$. Then the maximum value of $\sin \alpha$ is $\qquad$ .
2. $\frac{\sqrt{5}-1}{2}$.

Given that $\sin \alpha$ is a root of the equation, we have
$$
a \sin ^{2} \alpha+b \sin \alpha-b=0 \text {. }
$$

Obviously, $\sin \alpha \neq 0$, otherwise it would contradict $b>0$.
Also, from $a \geqslant b>0$, we get
$$
0=a \sin ^{2} \alpha+b \sin \alpha-b \geqslant b \sin ^{2} \alpha+b \sin \alpha-b \text {, }
$$

which means $\sin ^{2} \alpha+\sin \alpha-1 \leqslant 0$.
Solving this, we get $-1 \leqslant \sin \alpha \leqslant \frac{\sqrt{5}-1}{2}$.

When $a=b>0$, $\sin \alpha$ reaches its maximum value $\frac{\sqrt{5}-1}{2}$.