#### Answer

$256 \text{ feet}$

#### Work Step by Step

The variation model described by the problem is $
s=k\sqrt{d}
,$ where $
s
$ is the speed and $
d
$ is the distance.
Substituting the known values in the variation model above results to
\begin{array}{l}\require{cancel}
160=k\sqrt{400}
\\
160=k(20)
\\
\dfrac{160}{20}=k
\\
k=8
.\end{array}
Therefore, the variation equation is
\begin{array}{l}\require{cancel}
s=8\sqrt{d}
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
128=8\sqrt{d}
\\
\dfrac{128}{8}=\sqrt{d}
\\
16=\sqrt{d}
\\
(16)^2=(\sqrt{d})^2
\\
256=d
.\end{array}
Hence, the distance (or height of the cliff), $d,$ is $
256 \text{ feet}
.$