jmc

First, let us draw a diagram (not to scale!): Here, $AB$ is the telephone pole and $C$ is the point in the ground where the cable $BC$ is anchored. The key is to recognize that $ABC$ is a right triangle since the telephone pole is upright. Meanwhile, Leah stands at $D$ and touches the cable at $E,$ so $DEC$ is another right triangle. Not only that, but we see that $△ABC\sim △DEC$ thanks to AA similarity.

From the problem, We have that $DE=1.5\text{m},$ $AC=3\text{m},$ and $AD=2.5\text{m}.$ Therefore, $DC=AC-AD=0.5\text{m}.$ We desire $AB.$ From $△ABC\sim △DEC,$ we get:

$$\begin{array}{rl}\frac{AB}{AC}& =\frac{DE}{DC}\\ \frac{AB}{3\text{m}}& =\frac{1.5\text{m}}{0.5\text{m}}=3\\ AB& =3\cdot 3\text{m}=\boxed{9}\text{ meters}.\end{array}$$