= Solution
Take upward as positive and interpret the stated $97\,\mathrm{kg}$ as the total mass immediately before launch, including the harpoon. The remaining probe mass is $M=95.6\,\mathrm{kg}$. <Conservation of momentum> during the short firing interval gives
$$
-97(1)=Mv_+ -1.4(70),\qquad\boxed{v_+=\frac{1}{95.6}\,\mathrm{m\,s^{-1}}=0.01046\,\mathrm{m\,s^{-1}}.}
$$
The harpoon speed used here is relative to the comet, as printed. From $v_{\rm esc}^2=2GM_c/R$, the near-surface <gravitational acceleration> is
$$
g_c=\frac{v_{\rm esc}^2}{2R}=4.84\times10^{-5}\,\mathrm{m\,s^{-2}}.
$$
For the <near-surface recoil height with constant downward thrust>, the upward motion has downward acceleration $a=g_c+F/M$. Therefore
$$
\boxed{H=\frac{v_+^2}{2(g_c+F/M)}=0.358\,\mathrm m.}
$$
Since $H/R\simeq1.8\times10^{-4}$, treating gravity as constant is consistent. This uses only the stated gravity and thrust after firing; a taut tether could reduce this free-flight height, but no tether length or elasticity is specified. If instead $97\,\mathrm{kg}$ meant the probe mass after firing, the pre-firing mass would be $98.4\,\mathrm{kg}$ and its momentum would leave the probe still moving downward, with no upward excursion. This illustrates <nearly cancelling recoil momenta in payload ejection>. The mass convention matters because the launch momenta nearly cancel.
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