= Solution
\b[Powered vertical flight.] Take upward <velocity> $v$ as positive, and let $v_e>0$ be the downward exhaust <speed> relative to the rocket. During $dt$ the rocket's <mass> changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame <velocity> $v-v_e$. Compare the <momentum> of the rocket plus this newly expelled exhaust before and after the interval:
$$
(m+dm)(v+dv)+(-dm)(v-v_e)-mv=-mg\,dt.
$$
Discard the second-order product $dm\,dv$. The result is $m\,dv+v_e\,dm=-mg\,dt$, or
$$
m\frac{dv}{dt}=-v_e\frac{dm}{dt}-mg.
$$
The positive <rocket thrust> is $-v_e\dot m$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the <rocket equation> with gravitational loss:
$$
\boxed{v(t)=v_e\ln\frac{m_0}{m(t)}-gt.}
$$
Using $d(mv)/dt$ for the rocket alone without the exhaust <momentum flux> would give the wrong equation for this open system.
\b[Orbital speed under the stated gravity assumption.] A <circular orbit> of radius $r_o=R_E+h_o=7.90\times10^6\,\mathrm m$ requires <centripetal acceleration> $v_o^2/r_o$. If the magnitude of <gravitational acceleration> is the stipulated constant $g=9.8\,\mathrm{m\,s^{-2}}$ even at the orbit,
$$
\frac{v_o^2}{r_o}=g,
\qquad
\boxed{v_o=\sqrt{gr_o}=8.80\times10^3\,\mathrm{m\,s^{-1}}.}
$$
This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.
\b[Burnout height.] During the unpowered <projectile motion>, horizontal <velocity> is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital <speed> $v_o$. At burnout, therefore,
$$
v_{y,b}=v_o\tan30^\circ,\qquad
h_o-h_b=\frac{v_{y,b}^2}{2g}=\frac{v_o^2}{6g}.
$$
These follow either by integrating $\ddot y=-g$ or by the vertical <work-energy theorem>. Substituting the preceding constant-$g$ orbital result gives
$$
\boxed{h_b=h_o-\frac{R_E+h_o}{6}
=1.83\times10^5\,\mathrm m\simeq183\,\mathrm{km}.}
$$
The vertical component is initially positive, as required to reach the orbit after burnout.
\b[Alternative inverse-square interpretation.] A real Earth's exterior <Newtonian gravitational field> is not constant over a $1500\,\mathrm{km}$ altitude change. If $9.8\,\mathrm{m\,s^{-2}}$ instead means the surface value $g_0$, then <Newton's law of universal gravitation> gives $GM=g_0R_E^2$ and
$$
v_o=\sqrt{\frac{g_0R_E^2}{R_E+h_o}}=7.13\times10^3\,\mathrm{m\,s^{-1}}.
$$
Using this orbital <speed> but the paper's constant-$g$ flat <projectile motion> approximation for the coast would give
$$
\boxed{h_b=1.50\times10^6-
\frac{R_E^2}{6(R_E+h_o)}=6.36\times10^5\,\mathrm m\simeq636\,\mathrm{km}.}
$$
This alternative changes the global constant-gravity assumption; it is not the same model as the $183\,\mathrm{km}$ result. An exact curved <Kepler orbit> coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.
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