Question
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
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