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Let a massless bucket approach a wall under constant tension $T>0$, with remaining distance $x$ and remaining sand mass $m=Mx/L$. Assume zero-relative-velocity mass shedding horizontally. Then $m\ddot{x}=-T$ and, from rest at $x=L$, the speed satisfies $u^2=(2TL/M)\ln (L/x)$. Its kinetic energy is $K=Tx\ln (L/x)$ and its momentum magnitude is $p=x\sqrt{(2MT/L)\ln (L/x)}$. Differentiating gives $x_K=L/e$ and $x_p=L/\sqrt{e}$: the kinetic energy peaks later, closer to the wall, because its dependence on the increasing speed is stronger. Both tend to zero as the remaining mass vanishes, despite divergent ideal speed.

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 byThe leading external impulse is $(T-qg)dt$. Newton's second law applied to this combined material system therefore givesWriting $\frac{d}{dt}(qv)=T-qg$ 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. ThenIntegrating from rest,Since $A/t_0=1+M/m_0$ and $A/(A-t_0)=1+m_0/M$, the final upward speed isThe acceleration is nonnegative throughout release, consistent with the upward motion.