A rock is thrown straight up from a cliff edge with an initial velocity of 13 m/s. Taking up as positive and using g = 9.80 m/s^2, what is its position at t = 2.00 s?
A-5.10 m
B3.20 m
C6.40 m
D8.10 m
Answer & Solution
Correct answer: C. 6.40 m
1. Use y = y_0 + v_0 t - 0.5 g t^2 with y_0 = 0 and v_0 = 13 m/s.
2. y = (13.0 m/s)(2.00 s) - 0.5(9.80 m/s^2)(2.00 s)^2.
3. y = 26.0 m - 19.6 m = 6.40 m.
4. The positive value means the rock is still above the cliff edge at that moment.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_
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