alleryStart learning →The same rock is thrown straight up from a cliff edge at 13 m/s. Taking up as positive and using g = 9.80 m/s^2, what is its velocity at t = 3.00 s?
A-16.4 m/s
B-6.60 m/s
C3.20 m/s
D13.0 m/s
Answer & Solution
Correct answer: A. -16.4 m/s
1. Use v = v_0 - gt with v_0 = 13.0 m/s and g = 9.80 m/s^2.
2. v = 13.0 m/s - (9.80 m/s^2)(3.00 s).
3. v = 13.0 m/s - 29.4 m/s = -16.4 m/s.
4. The negative sign shows the rock is now falling, well past the top of its path.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_
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