= Solution
The specified <rocket fuel fraction> gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yields
$$
T=\frac{u\log(3/2)-v_0}{g}.
$$
Then $h(T)=0$ fixes
$$
h_0=(v_0-u)T+\frac12gT^2+2uT\log(3/2).
$$
Numerically,
$$
\boxed{T\simeq60.07\,\mathrm s,\qquad h_0\simeq1.159\times10^3\,\mathrm m.}
$$
The associated constant <mass loss rate> is $\alpha=m_0/(3T)\simeq88.78\,\mathrm{kg\,s^{-1}}$. The initial upward <thrust acceleration> already exceeds $g$ and increases as <mass> falls, so the downward <speed> decreases monotonically to zero. This calculation assumes constant lunar <gravitational acceleration>, constant relative <rocket exhaust speed>, and negligible <drag>.
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