alleryStart learning →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_
Keep practising Acceleration
Browse all Acceleration questions for NCEA Level 1, or explore more Physics topics.
Related questions
Drag racers compete over a standing start on a track of what length?During launch and reentry, about how many g's does the space shuttle produce for its occupFor an object falling under constant gravity, which pair of graphs against time are straigUnder what condition is a velocity versus time graph a straight line rather than a curve?If a plot of displacement versus time is a straight line, what does that say about the objHow long does it take to accelerate from 8.0 m/s to 20.0 m/s at a constant rate of 3.0 m/sA velocity increases from 0 to 20 m/s over 10 s at a constant rate. What is the average acThe same rock is thrown straight up from a cliff edge at 13 m/s. Taking up as positive and