The Vertical Launch of a Rocket

As you say :

$$M(t) = M_o - \alpha t$$

But you must know that at time $\tau$ after start the object now has $0.1M_o$ mass (since it consumed all of it's fule).

Therfore

$$0.1M_o = M_o - \alpha \tau \tag 1$$ $$\Rightarrow \tau = \frac {0.9M_o}{\alpha} \tag 2$$

After substituting $(1)$ and $(2)$ into your equation we get:

$$ v(\tau) = u\ln \left(10 \right)-g\frac {0.9M_o}{\alpha}$$

Now if you know $\alpha$ then you can find $v(\tau)$.

Based on a very nice answer from @Johan Liebert, you can extrapolate several things. First, upper bound of rocket speed is when $\alpha \to \infty$, this gives : $$ v_{max} = u\,\ln(10) $$, because second term approaches zero then.

Second - you can calculate critical fuel consumption, solving for $\alpha$ in : $$ u\,\ln(10) - g \frac{0.9 M_o}{\alpha} = 0 $$ this gives : $$ \alpha_{\text{critical}} = g\frac{0.9\,M_o}{u \ln(10)} $$ If $\alpha < \alpha_{\text{critical}}$, then rocket finally will start to fallback to Earth, i.e. gravity force will win in the end.