## Intermediate Algebra (6th Edition)

$256 \text{ feet}$
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 $256 \text{ feet} .$