Answer
Just before the skier lands, her speed is $10.9m/s$
Work Step by Step
1) As we can see in the figure below, the component of the skier's weight $mg\sin25$ helps to propel the skier down. Kinetic friction opposes this motion. So, on the slope, $\sum F=mg\sin25-f_k=0.423mg-f_k$
We know $f_k=\mu_kF_N$. Again, from the figure, we see that $F_N$ is in the opposite direction with $mg\cos25$. As there is no vertical acceleration, $F_N=mg\cos25=0.906mg$. Therefore, $f_k=0.906\mu_kmg$
In short, $$\sum F=0.423mg-0.906\mu_kmg$$
We know $\mu_k=0.2$ and $g=9.8m/s^2$, so $$\sum F=4.15m-1.78m=2.37m$$
The work done by $\sum F$ over the slope with $s=10.4m$ on the skier ($\theta=0^o$ since the force is parallel with the motion) is $$W=(\sum F\cos0)s=24.65m (1)$$
From the work-energy theorem, $$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2$$
The skier starts from rest, $v_0=0$, so $$W=\frac{1}{2}mv_f^2$$
Plug (1) in here, $$24.65m=\frac{1}{2}mv_f^2$$ $$v_f=\sqrt{24.65\times2}=7.02m/s$$
which is the speed of the skier just before the edge of the cliff.
2) Now we take the speed of the skier just before the edge of the cliff to be $v_0=7.02m/s$ and use equations of kinematics to find $v_f$
When the skier skies off the cliff, she free falls, so $g=9.8m/s^2$. We also know $s=3.5m$. Therefore, $$v_f^2=v_0^2+2gs=117.88$$ $$v_f=10.9m/s$$
