#### Answer

(a) The skier's velocity at the bottom of the hill is 15.9 m/s
(b) The skier can travel 30.5 meters up the slope.

#### Work Step by Step

(a) The acceleration directed down the incline is $g~sin(\theta)$. Therefore;
$v^2 = v_0^2+2ad = 0 + 2ad$
$v = \sqrt{2ad} = \sqrt{(2)(9.80~m/s^2)~sin(15^{\circ})(50~m)}$
$v = 15.9~m/s$
The skier's velocity at the bottom of the hill is 15.9 m/s
(b) The acceleration directed down the slope is $g~sin(\theta)$. So;
$d = \frac{v^2-v_0^2}{2a}$
$d = \frac{0-(15.9~m/s)^2}{(2)(-9.80~m/s^2)~sin(25^{\circ})}$
$d = 30.5~m$
The skier can travel 30.5 meters up the slope.