Answer
$v_0=4.54m/s$
Work Step by Step
According to the work-energy theorem, $$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2$$
We do not know the mass of the skier $m$ and his initial speed $v_0$. But we know $v_f=0$, so we can rewrite the equation: $$W=-\frac{1}{2}mv_0^2 (1)$$
Kinetic friction $f_k$ acts as the only force influencing the skier's motion by stopping it. $f_k$ opposes the motion in the opposite direction of the displacement, so $$W=(f_k\cos180)s=-f_ks$$
We have $f_k=\mu_kF_N=\mu_kmg$ (because there is no vertical acceleration). So, $$W=-\mu_kmgs (2)$$
From (1) and (2), we have $$-\frac{1}{2}mv_0^2=-\mu_kmgs$$ $$\frac{1}{2}v_0^2=\mu_kgs$$ $$v_0=\sqrt{2\mu_kgs}$$
We have $\mu_k=0.05$, $g=9.8m/s^2$ and $s=21m$
$$v_0=4.54m/s$$
