= Solution
During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward <momentum>, including gravity's impulse and exhaust <velocity> $v-U$, gives to first order
$$
m\,dv+U\,dm=-mg\,dt,
$$
which is the stated rocket equation.
Here $dm/dt=-\alpha$ and $U=U_0m_0/m$, so
$$
\frac{dv}{dt}=\frac{\alpha U_0m_0}{(m_0-\alpha t)^2}-g.
$$
Lift-off from rest requires positive initial <acceleration>,
$$
\frac{\alpha U_0}{m_0}>g.
$$
With $v(0)=0$, integration gives
$$
v(t)=U_0\left(\frac{m_0}{m_0-\alpha t}-1\right)-gt.
$$
The dimensions are $[m_0]=[m]=M$, $[\alpha]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.
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