= Solution
Use <zero-relative-velocity mass shedding>: the released sand initially shares the balloon's vertical <velocity>, rather than being actively ejected. Let $q=M+m(t)$ and let a positive mass $\delta m$ be shed in a short interval $dt$. The <momentum> of the balloon and that newly released sand changes by
$$
(q-\delta m)(v+dv)+\delta m\,v-qv=q\,dv+O(dt^2).
$$
The leading external impulse is $(T-qg)dt$. <Newton's second law> applied to this combined material system therefore gives
$$
\boxed{(M+m)\dot v=T-(M+m)g.}
$$
Writing $\frac d{dt}(qv)=T-qg$ for the remaining balloon alone would omit the outgoing <momentum> flux. A nonzero relative ejection <velocity> would instead produce a <rocket equation> thrust term.
Initial mechanical equilibrium gives $T=(M+m_0)g$. Write $\lambda=m_0/t_0$, so $m=m_0-\lambda t$ up to depletion. Then
$$
\dot v=g\frac{\lambda t}{M+m_0-\lambda t}
=g\frac{t}{A-t},\qquad A=\frac{M+m_0}{\lambda}>t_0.
$$
Integrating from rest,
$$
v(t)=g\left[A\log\frac{A}{A-t}-t\right].
$$
Since $A/t_0=1+M/m_0$ and $A/(A-t_0)=1+m_0/M$, the final upward <speed> is
$$
\boxed{v(t_0)=gt_0\left[\left(1+\frac{M}{m_0}\right)
\log\left(1+\frac{m_0}{M}\right)-1\right].}
$$
The acceleration is nonnegative throughout release, consistent with the upward motion.
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