Posted by Carter on .
A 75 kg skier starts at rest on top of a icy inclined plane (with a 30 degree angle to the ground). He ends up going past the inclined plane and onto a flat ground. And comes to a rest. The coefficient of friction is .100. How long does it take for the skier to go from rest (at the top of the inclined plane) to rest (on the flat ground)?
There is no more information available. I need help solving it.

Physics 
Henry,
We need to know the length of the hill.
Length = 100 m?
A = 30o
m = 75 kg
h = 100*sin30 = 50 m. = Ht. of hill.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*50 = 980
V = 31.3 m/s At bottom of hill.
V^2 = Vo^2 + 2a*d
a=(V^2Vo^2)/2d=(31.3^20)/200=4.9 m/s^2
V = Vo + a*t = 31.3
0 + 4.9*t = 31.3
T1 = 6.39 s. = Time to reach gnd.
V = Vo + a*t = 0
31.3  4.9t = 0
4.9t = 31.3
T2 = 6.39 s. on level gnd.
T = T1 + T2 = 6.39 + 6.39 = 12.78 s.