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ussjt
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An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 13.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.20 m below the edge. How fast is she going just before she lands?
W= Fn*13.4
Fn*13.4= .5mv^2
(mg sin25 - (.200)(mg cos25))(13.4)= .5mv^2
(g sin25 - (.200)(g cos25))(13.4)= .5v^2
((9.8) sin25 - (.200)((9.8) cos25))(13.4)= .5v^2
31.6949624131 = .5v^2
2(31.6949624131) = v^2
sqrt (63.3899248261) = v
7.96177899882 = v
~~~~~~~~~~~~~~~~~~~~~~~~~~
Vy= 7.96177899882 (sin25)
Vy= 3.36479320085
~~~~~~~~~~~~~~~~~~~~~~~~~~
Vy = sqrt (3.36479320085^2 + 2(-9.8)(-3.2))
Vy = 8.604756 m/s
W= Fn*13.4
Fn*13.4= .5mv^2
(mg sin25 - (.200)(mg cos25))(13.4)= .5mv^2
(g sin25 - (.200)(g cos25))(13.4)= .5v^2
((9.8) sin25 - (.200)((9.8) cos25))(13.4)= .5v^2
31.6949624131 = .5v^2
2(31.6949624131) = v^2
sqrt (63.3899248261) = v
7.96177899882 = v
~~~~~~~~~~~~~~~~~~~~~~~~~~
Vy= 7.96177899882 (sin25)
Vy= 3.36479320085
~~~~~~~~~~~~~~~~~~~~~~~~~~
Vy = sqrt (3.36479320085^2 + 2(-9.8)(-3.2))
Vy = 8.604756 m/s