Physics- need help!!?

Question

If you were on a 26.9 degree slope and an avalanche started 450 m up the slope, how much time would you have to get out of the way? Answer in s.

Answer 1

Assume the slope is frictionless (quite right on the icy surface). and let the mass of the avalanche to be M.
The weight of the avalanche is Mg.
The component parallel to the surface of the slope, and would cause movement along the slope, is Mg*sin(26.9°). According to Newton's third Law F= ma: we have:
Mg*sin(26.9°) = Ma
Hence a = g*sin(26.9°)
Since 0.5*a*t^2 = S = 450m
we find the time to be:
t = sqrt(2*s/a) = sqrt(900/(g*sin(26.9°))) = 14.2 (s)
within which you have to get out of the way.