= Solution
The exact no-penetration condition on the stationary boundary is that the velocity be tangent to it. Linearized at $z=0$, it is
$$
w'(x,0)=U\frac{dh}{dx}
=-Uk h_0\sin kx.
$$
The terrain is stationary, so $\omega=0$ and $\widehat\omega=-kU$. The dispersion relation gives
$$
m^2=\frac{N^2}{U^2}-k^2.
$$
A vertically propagating wave therefore exists if and only if
$$
|kU|<N.
$$
The upward <radiation condition> $\widehat\omega m<0$ selects $km>0$ for $U>0$. Taking $k>0$ gives $m>0$. Lines of constant phase satisfy
$$
kx+mz=\text{constant},
$$
so they tilt upstream as height increases. The discarded sign of $m$ gives the mirror-image phase tilt and downward energy propagation.
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