= Solution
Assume constant <mass density>, a prismatic channel, approximately uniform cross-sectional <velocity> $u$, a <hydrostatic pressure> distribution, small bed slope $S$, and no lateral inflow. Integrating <mass conservation> over a slice of length $dx$ gives $\rho A_tdx=\rho Au|_x-\rho Au|_{x+dx}$, hence $A_t+(Au)_x=0$.
Let $I(A)=\int_A(H-z)\,dA$ be the hydrostatic first moment, where $H$ is the actual free-surface depth. The downstream <momentum> balance is
$$
(Au)_t+(Au^2+gI)_x=gAS-\frac{\tau\ell}{\rho}.
$$
Using $A_t+(Au)_x=0$ reduces its left side to $A(u_t+uu_x)+gI_x$. For a fixed prismatic cross-section, $dI/dH=A$, so $I_x/A=H_x$. Therefore the exact depth-averaged result is
$$
\boxed{u_t+uu_x=gS-\frac{\tau}{\rho R}-gH_x,\qquad R=A/\ell.}
$$
Here <hydraulic radius> converts perimeter <shear stress> into <force> per unit fluid volume. The printed notation $\bar h$ yields the requested equation if it denotes this hydraulic <pressure> head, whose <derivative> equals $H_x$.
\b[The geometrical definition printed in the next part needs care.] <Triangular-channel hydrostatic pressure head> gives the factor explicitly. For the stated symmetric triangular notch, the actual depth is $H=(\ell/2)\sin\theta$, whereas its printed $\bar h=(\ell/4)\sin\theta$ is $H/2$. Indeed $A=H^2\cot\theta$ and $I=AH/3$, giving $I_x/A=H_x=2\bar h_x$. Thus the equation with precisely the printed geometrical $\bar h$ has <pressure> term $-2g\bar h_x$. A first-principles derivation cannot remove that factor. The subsequent intended scaling can be carried out algebraically using the printed convention, as below, or with the <pressure> scale doubled for the actual triangular channel.
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