Answer
Santa's speed as he leaves the roof is 9.9 m/s
Work Step by Step
The acceleration directed down the incline is $g~sin(\theta)$. Therefore;
$v^2 = v_0^2+2ad = 0 + 2ad$
$v = \sqrt{2ad} = \sqrt{(2)(9.80~m/s^2)~sin(30^{\circ})(10~m)}$
$v = 9.9~m/s$
Santa's speed as he leaves the roof is 9.9 m/s
