 Solution

Newton's first law identifies inertial frames in which a force-free body maintains constant velocity. The second is $\dot{\mathbf{P}}={\mathbf{F}}_{\mathrm{net}}$ for a closed system; for fixed mass it becomes $m\mathbf{a}=\mathbf{F}$. The third gives equal and opposite pair forces. Summing the second law over a closed collection cancels internal force pairs, so total momentum is conserved when the resultant external force vanishes. A rocket alone is an open variable-mass subsystem, so one must include the momentum carried by exhaust before applying this reasoning.

Choose downward velocity as positive and write $M(t)=M_0-\alpha t$. During $dt$, exhaust mass $dm=\alpha dt$ leaves downward with speed $v+v_E$ in the surface frame. To first order, momentum balance for rocket plus this exhaust givesIntegration, with $v(0)=v_0$, yieldsSetting $g=0$ gives the gravity-free result. This assumes a constant downward relative exhaust speed, constant mass-loss rate and $M_0-\alpha t>0$ throughout the burn; the negative logarithm supplies the upward thrust effect.

Height satisfies $\dot{h}=-v$. Put $u(t)=1-\alpha t/M_0$. Integrating the logarithm and choosing $h(0)=h_0$ givesEquivalently, the last term is $v_E[t+(M_0-\alpha t)\log u/\alpha ]$. Differentiating either form recovers $-v$ and checks the sign.

For this constant-rate retro-rocket landing, at touchdown $u_L=3/4$ and $v_L=0$, henceThe fuel condition gives $\alpha =M_0/(4t_L)=200.30$ kg/s. Requiring $h(t_L)=0$ then givesIn this burn the downward speed decreases monotonically to zero: the initial thrust acceleration already exceeds gravity and increases as fuel is consumed. The result is the idealized vertical landing, without atmospheric drag, changing gravity or a separate terminal burn phase.