## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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