The same rock is thrown straight up at 13.0 m/s with upward taken as positive. What is its velocity 1.00 s after release?
A-3.20 m/s, so it is already falling
B+3.20 m/s, so it is still climbing
C+13.0 m/s, so its speed is unchanged
D+9.80 m/s, so it is still climbing
Answer & Solution
Correct answer: B. +3.20 m/s, so it is still climbing
1. Identify the knowns: the initial velocity is +13.0 m/s, the acceleration is -9.80 m/s^2 and the time is 1.00 s.
2. Use final velocity equals initial velocity minus the acceleration due to gravity times time.
3. Substitute the values: 13.0 m/s - (9.80 m/s^2)(1.00 s).
4. This gives 13.0 m/s - 9.80 m/s = +3.20 m/s.
5. The positive sign means the rock is still heading upward at that instant.
6. It has slowed from its original 13.0 m/s, exactly as expected.
7. Reading the answer as -3.20 m/s would place the rock on its way back down too early.
_Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.7 Falling Objects_
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