A Skier starting from rest rolls down a 35.0 m ramp. When he arrives at the bottom of the ramp his speed is 10.43 m/s. (a) Determine the magnitude of his acceleration, assumed to be constant, (b) If the ramp is inclined at 60.0 degrees with respect to the ground, what is the component of his acceleration that is parallel to the ground. (c) Provide me with a free body diagram, one for the 35 m ramp situation. Also, provide a nice sketch of the skier.

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-07-15T19:32:24+02:00

We obtain the magnitude of the acceleration as 1.55 m/s^2 while its component parallel to the ground is 0.775 m/s^2.

The acceleration refers to change of speed with time. From the question;

It the follows that;

v^2 = u^2 + 2as

And;

u = 0m/s owing to the fact that motion started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

For the acceleration component which is parallel to the ground;

ax = a sin θ

ax = 1.55 m/s^2 cos 60

ax = 0.775 m/s^2

Learn more about acceleration: brainly.com/question/12550364

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