Askier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at the bottom is 15 m/s. what is the lenght of the incline? how long does it take him to reach the bottom?

... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s

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