#### Answer

The skier's speed at the bottom is 15.8 m/s

#### Work Step by Step

We can find the distance $d$ that the skier travels down the slope.
$\frac{h}{d} = sin(\theta)$
$d = \frac{h}{sin(\theta)}$
$d = \frac{50~m}{sin(20^{\circ})}$
$d = 146~m$
The skier's kinetic energy at the bottom is equal to the sum of the initial potential energy and the work $W_w$ done on the skier by the wind.
$KE = PE+W_w$
$\frac{1}{2}mv^2 = mgh+F~d~cos(160^{\circ})$
$v^2 = \frac{2mgh+2F~d~cos(160^{\circ})}{m}$
$v = \sqrt{\frac{2mgh+2F~d~cos(160^{\circ})}{m}}$
$v = \sqrt{\frac{(2)(75~kg)(9.80~m/s^2)(50~m)+(2)(200~N)(146~m)~cos(160^{\circ})}{75~kg}}$
$v = 15.8~m/s$
The skier's speed at the bottom is 15.8 m/s