Asked by Emmi
A skier reaches a speed of 42 m/s on a 25◦ski slope. Ignoring friction, what was the distance along the slope the skier would have had to travel, starting from rest?
Answers
Answered by
Anonymous
component of weight force down slope = m g sin 25
so
acceleration a = 9.81 sin 25 = 4.15 m/s^2
v = 4.15 t
42 = 4.15 t
t = 10.1 seconds
d = (1/2) a t^2
d = (1/2) *4.15 *10.1^2 = 840 meters
so
acceleration a = 9.81 sin 25 = 4.15 m/s^2
v = 4.15 t
42 = 4.15 t
t = 10.1 seconds
d = (1/2) a t^2
d = (1/2) *4.15 *10.1^2 = 840 meters
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