- #1
punyhuman92
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1. A skier starts from rest at the top of a hill of height h down a 30.0 degree ski slope with an acceleration a.
a.) What is her speed at the bottom of the hill?
b.) If the hill was h = 250. m high and her acceleration down the slope was a = 4.00 m/s2, what was instantaneous speed at the bottom of the hill, and her average speed down the slope?
c.) At the base of the hill, she continues horizontally for another 250 m before coming to a stop and ending her run. What is her total vector displacement from start to finish?
I'm asking about a) for now.
for a.) I drew the FBD, rotated it so Fgx is parallel to the slope of the hill.
mgSin30=Fgx
Fgx=4.9m
then I did Fnet/m=a
the m's cancel so I'm left with a=4.9 m/s^2
it's asking for the speed though, aka velocity so is my answer expected to simply be 4.9m/s^2*t? I feel like I should be able to work with it more but i don't see how I could?