= Solution
At distance $x$ below the support, the cable must carry the weight of the remaining length $L-x$. Thus its equilibrium <tension> is $T(x)=\mu g(L-x)$, and the local <transverse wave> speed is
$$
\boxed{c(x)=\sqrt{\frac{T(x)}\mu}=\sqrt{g(L-x)},\qquad0\leq x<L.}
$$
The <linear mass density> cancels. More precisely the small-displacement equation with variable <tension> is $\mu y_{tt}=\partial_x[T(x)y_x]$; its local characteristic speed is the expression above, even though a first-spatial-derivative term is also present.
The <wave travel time on a hanging uniform cable> is obtained from $dt=dx/c(x)$:
$$
t_{\mathrm{down}}=\int_0^L\frac{dx}{\sqrt{g(L-x)}}
=\frac1{\sqrt g}\int_0^Lu^{-1/2}\,du
=\boxed{2\sqrt{\frac Lg}.}
$$
Although the local speed tends to zero at the free end, the endpoint singularity in the travel-time integral is integrable. This finite result belongs to the ideal flexible-cable model, with bending stiffness neglected.
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