Answer
The firefighter slid $5.3m$ down the pole.
Work Step by Step
According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=W_{f}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{f}$$
First, we notice that his initial speed $v_0=0$ and his final position has $h_f=0$, so we can simplify the equation $$\frac{1}{2}mv_f^2-mgh_0=W_f$$
We have $m=93kg$ and $v_f=3.4m/s$
The work done by frictional force $W_f=(f\cos\theta) s=-fs$ (since the force opposes the motion). $f=810N$ and $s=|h_f-h_0|=h_0$, so $W_f=-810h_0$
Plug these back to the equation: $$537.54-911.4h_0=-810h_0$$ $$101.4h_0=537.54$$ $$h_0=5.3m$$
