= Solution
\b[Powered ascent.] Use upward signed <velocity>, constant <gravitational acceleration> $g=9.81\,\mathrm{m\,s^{-2}}$, constant relative <rocket exhaust speed> $u=2500\,\mathrm{m\,s^{-1}}$, vertical flight from rest, no <drag force>, and no <impulse> at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying <Newtonian gravitational field> would require integrating a different ascent equation. Put $m_0=540000\,\mathrm{kg}$ and $q=2500\,\mathrm{kg\,s^{-1}}$. During ascent $m(t)=m_0-qt$, and <momentum> balance gives the <rocket equation> with <gravity>:
$$
m\dot v=uq-mg,
\qquad v(t)=u\ln\frac{m_0}{m_0-qt}-gt.
$$
The launch <rocket thrust> $uq=6.25\times10^6\,\mathrm N$ exceeds $m_0g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has <mass> $m_b=140000\,\mathrm{kg}$, not $40000\,\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint <speed> are
$$
t_b=\frac{400000}{2500}=160\,\mathrm s,
\qquad
\boxed{v_b=u\ln\frac{m_0}{m_b}-gt_b
=1805.22\,\mathrm{m\,s^{-1}}.}
$$
\b[Coasting height.] Integrating the ascent <velocity> from zero height gives
$$
z(t)=u\left[t-\frac{m_0-qt}{q}\ln\frac{m_0}{m_0-qt}\right]-\frac12gt^2.
$$
For example this follows by substituting $m=m_0-qt$ in $\int_0^t\ln[m_0/(m_0-qs)]\,ds$. At separation,
$$
z_b=u\left[t_b-\frac{m_b}{q}\ln\frac{m_0}{m_b}\right]-\frac12gt_b^2
=85442.26\,\mathrm m.
$$
The separated stage then undergoes <free fall in a constant gravitational field> with initial upward <speed> $v_b$. <Conservation of mechanical energy> during this coast gives
$$
\boxed{H=z_b+\frac{v_b^2}{2g}=251538.47\,\mathrm m\simeq251.54\,\mathrm{km}.}
$$
Its smaller <mass> after separation does not change that coasting height.
\b[Reserve burn.] On descending to $z_i=100000\,\mathrm m$, <conservation of mechanical energy> gives downward <speed>
$$
w_i=\sqrt{2g(H-z_i)}=1724.29255\,\mathrm{m\,s^{-1}}.
$$
Reset time to $s=0$ at reserve ignition. Now $m_i=40000\,\mathrm{kg}$, $m_f=14000\,\mathrm{kg}$ and $q_r=26000/92\,\mathrm{kg\,s^{-1}}$. Downward ejection supplies upward <rocket thrust>. The upward signed <velocity> is
$$
v_r(s)=-w_i+u\ln\frac{m_i}{m_i-q_rs}-gs.
$$
Thus the formal free-flight burnout value is
$$
\boxed{v_r(92)=-w_i+u\ln\frac{40000}{14000}-92g
=-2.25724\,\mathrm{m\,s^{-1}},}
$$
corresponding to a \b[formal downward speed of $2.26\,\mathrm{m\,s^{-1}}$]. Keeping $g=9.8\,\mathrm{m\,s^{-2}}$ throughout instead gives $3.82\,\mathrm{m\,s^{-1}}$ downward; premature rounding is especially damaging because three large <velocity> terms almost cancel.
\b[The free-flight endpoint is below ground.] Integrating the same reserve-burn <velocity> gives
$$
z_r(s)=z_i-w_is+u\left[s-\frac{m_i-q_rs}{q_r}\ln\frac{m_i}{m_i-q_rs}\right]-\frac12gs^2.
$$
At the stated endpoint this gives $z_r(92)=-167.27\,\mathrm m$ for $g=9.81\,\mathrm{m\,s^{-2}}$. The surface is crossed at approximately $s=89.15\,\mathrm s$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout <speed> cannot be inferred without a ground-contact model. The boxed <velocity> is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The <constant-rate retro-rocket landing> must satisfy both its <velocity> and height conditions.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2021-resit-reserve-burn.png]
{title=Constant-gravity reserve descent: the formal trajectory crosses ground before its nominal burnout}
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