A rock thrown straight up from a cliff edge is at +6.40 m with a velocity of -6.60 m/s at 2.00 s. What does this describe?
ABelow the release point and still moving upward
BAbove the release point and still moving upward
CAbove the release point but now moving downward
DBelow the release point and now moving downward
Answer & Solution
Correct answer: C. Above the release point but now moving downward
1. Upward has been taken as positive, so signs carry the directions.
2. A position of +6.40 m is above the release point, since it is positive.
3. A velocity of -6.60 m/s points downward, since it is negative.
4. The rock is therefore above where it started but already falling back.
5. Position and velocity are independent here, so the sign of one does not fix the sign of the other.
_Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.7 Falling Objects_
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