= Solution
Manning's uniform-flow law implies the friction slope $S_f=n^2u^2/R^{4/3}$ for positive downstream flow. In a wide rectangular section $A\simeq wh$ and $R\simeq h$, so $\tau l/(\rho A)=g n^2u^2/h^{4/3}$.
Scale distance with $L$, depth with $h_0$, <velocity> with $u_0=h_0^{2/3}S^{1/2}/n$, and time with $L/u_0$. Divide momentum by $gS$ to obtain
$$
\boxed{\epsilon=\frac{h_0}{SL},\qquad F=\frac{u_0}{\sqrt{gh_0}},\qquad E=\frac{LM}{wh_0u_0},}
$$
$$
\boxed{h_t+(hu)_s=E,\qquad\epsilon F^2(u_t+uu_s)=1-u^2/h^{4/3}-\epsilon h_s.}
$$
The factor $\epsilon$ compares depth change over $L$ with the mean bed drop $SL$, rather than with $L$ alone. Its <product> with $F^2$ measures inertial acceleration relative to bed-slope forcing.
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