= Solution
At the <grounding line>, hydrostatic flotation and continuity of ice flux give
$$
\boxed{h(x_G)=h_G=\frac{\rho_w}{\rho}b,
\qquad q(x_G^-)=q(x_G^+).}
$$
The normal stress and longitudinal membrane stress also match onto the floating <ice shelf>. For Newtonian ice this matching supplies the <Newtonian grounding-line flux>
$$
\boxed{q_G=\frac{\sqrt{gg'}}{6\nu}h_G^3,
\qquad
g'=g\frac{\rho_w-\rho}{\rho_w}.}
$$
The accumulation region still has $q=Ax$. In the ablation region, $q=A(2x_s-x)$, so $q(x_G)=q_G$ gives
$$
\boxed{x_G=2x_s-\frac{\sqrt{gg'}}{6\nu A}h_G^3.}
$$
Integrating the grounded flux law from $x_s$ to $x_G$ gives
$$
x_s^2=\frac g{6\nu A}(h_s^4-h_G^4)
+\left(\frac{q_G}{A}\right)^2.
$$
Under the stated limit $A\ll g'h_G^2/\nu$, the $h_G^4$ term is asymptotically smaller than the grounding-flux term, and hence
$$
\boxed{x_s\sim\left(\frac g{6\nu A}\right)^{1/2}
\left(h_s^4+\frac{g'}{6\nu A}h_G^6\right)^{1/2}.}
$$
Finally the accumulation-region profile gives $h_0^4=h_s^4+6\nu Ax_s^2/g$, so
$$
\boxed{h_0\sim\left(2h_s^4+\frac{g'}{6\nu A}h_G^6\right)^{1/4}.}
$$
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