A 75.0 kg novice skier is going down a hill with several secondary hills. Ignore frictional effects
and assume the skier does not push off or slow himself down. If the initial hill is 38.7 m above the
chalet level: (8 marks)
a) What is the speed of the skier at the bottom of the first hill, 25.6 m above the chalet level?
b) At what height does the skier have a speed of 12.0 m/s?
c) What is the skier's speed at the bottom of the hill?
d) Is this a realistic scenario?
2 answers
I'm not sure what to do
loss of potential energy = gain in kinetic energy
from 38.7 to 25.6:
m g (38.7 - 25.6) = (1/2) m v^2
v = sqrt (2 * 9.81 * 13.1)
= 16 m/s
12 = sqrt (2 * 9.81 * distance down)
144 = 19.62 d
d = 7.34 distance of fall
so 38.7 - 7.34 = 31.3 meters high at 12 m/s
v = sqrt (2 *9.81* 38.7
= 27.6 m/s
that is about 100 km/hr or 60 miles/hr
drag of snow and air would prevent this, hopefully :)
from 38.7 to 25.6:
m g (38.7 - 25.6) = (1/2) m v^2
v = sqrt (2 * 9.81 * 13.1)
= 16 m/s
12 = sqrt (2 * 9.81 * distance down)
144 = 19.62 d
d = 7.34 distance of fall
so 38.7 - 7.34 = 31.3 meters high at 12 m/s
v = sqrt (2 *9.81* 38.7
= 27.6 m/s
that is about 100 km/hr or 60 miles/hr
drag of snow and air would prevent this, hopefully :)
0 answers