posted by carol
A novice skier, starting from rest, slides down a frictionless 33.0° incline whose vertical height is 127 m. How fast is she going when she reaches the bottom?
The inclined plane forms a rt. triangle:
Y(ver) = 127 m,
A = 33 deg = angle bet. hor side and hyp.
d(hyp) = Y / sinA = 127 / sin33 = 233.2 m.
Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2 * 9.8 * 233.2,
Vf^2 = 4570.7,
Vf = 67.6 m/s = Final velocity.
V=sq.root(2gh)=sq.root-- 2 x 9.8m/s x 127m=49.89m/s
Angle does not really matter.