= Solution
For a hydrostatic <Boussinesq approximation>, introduce reduced pressure $\pi=p'/\rho_0$ and <buoyancy> $b=-g\rho'/\rho_0$. The <geostrophic balance> and perturbation <hydrostatic balance> equations are
$$
f u_g=-\pi_y,\qquad f v_g=\pi_x,\qquad\pi_z=b.
$$
Commuting the mixed derivatives gives the height-coordinate <thermal wind> relations
$$
\boxed{f\frac{\partial u_g}{\partial z}=-\frac{\partial b}{\partial y},\qquad
f\frac{\partial v_g}{\partial z}=\frac{\partial b}{\partial x}.}
$$
Equivalently $u_{g,z}=g\rho'_y/(f\rho_0)$ and $v_{g,z}=-g\rho'_x/(f\rho_0)$. In an atmosphere the corresponding ideal-gas form is conveniently derived from geopotential $\Phi$ on pressure surfaces: $fu_g=-\Phi_y|_p$, $fv_g=\Phi_x|_p$, and $\Phi_{\ln p}=-R_aT$. Hence
$$
f u_{g,\ln p}=R_aT_y|_p,\qquad f v_{g,\ln p}=-R_aT_x|_p.
$$
Since $d\ln p/dz=-g/(R_aT)$, this gives the vertical shear at a fixed horizontal location,
$$
\boxed{u_{g,z}=-\frac{g}{fT}T_y\big|_p,\qquad v_{g,z}=\frac{g}{fT}T_x\big|_p.}
$$
The horizontal temperature derivatives here are on isobars; replacing them by fixed-height derivatives is a small-slope approximation. This qualification avoids mixing compressible and Boussinesq density gradients.
For an illustrative poleward temperature drop $40\,\mathrm K$ across $3000\,\mathrm{km}$, mean $T\sim250\,\mathrm K$, $f\sim10^{-4}\,\mathrm{s^{-1}}$ and depth $10\,\mathrm{km}$,
$$
\boxed{\Delta u_g\sim\frac{gH\Delta T}{fT L}\simeq52\,\mathrm{m\,s^{-1}}.}
$$
Adding a modest surface wind gives an upper-level westerly jet of order $50$–$60\,\mathrm{m\,s^{-1}}$. The <thermal wind> fixes vertical shear, not the absolute wind without a boundary or reference value.
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