= Solution
Use the <gravity-current front condition> $\dot L=\operatorname{Fr}\sqrt{d\phi h}$, with a specified order-one front <Froude number>. In a <triangular-channel gravity-current box model>, the depth and concentration are uniform over $0<x<L(t)$. The volume is
$$
V=\frac\lambda2h^2L^2=\frac\lambda2H^2L_0^2,
\qquad \boxed{hL=HL_0=:K.}
$$
The horizontal projected deposition area is $\int_0^LT\,dx=\lambda hL^2$. Consequently the integral model is
$$
\boxed{\dot L=\operatorname{Fr}\sqrt{\frac{dK\phi}{L}},\qquad
\dot\phi=-\frac{2W_sL}{K}\phi,\qquad h=\frac KL,\qquad
L(0)=L_0,\quad\phi(0)=\phi_0.}
$$
This front speed is a closure for the integral model, rather than the assertion that its uniform interior is an exact solution of the spatially varying <shallow water equations>.
Eliminating time and integrating gives
$$
\frac{d\sqrt\phi}{dL}=-\frac{W_sL^{3/2}}{\operatorname{Fr}\sqrt d\,K^{3/2}},\qquad
\sqrt{\phi(L)}=\sqrt{\phi_0}-\frac{2W_s}{5\operatorname{Fr}\sqrt d\,K^{3/2}}(L^{5/2}-L_0^{5/2}).
$$
The limiting <runout length of a gravity current> is therefore
$$
\boxed{L_\infty=\left[L_0^{5/2}+\frac{5\operatorname{Fr}\sqrt{g'_0}\,(HL_0)^{3/2}}{2W_s}\right]^{2/5},\qquad g'_0=d\phi_0.}
$$
The additional distance beyond the release gate is $L_\infty-L_0$. The limit is approached asymptotically as $\phi\to0$ and $\dot L\to0$; the idealized box model does not predict a finite-time stop. Its numerical runout depends on the adopted front <Froude number>.
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