In an experiment two identical rocks are simultaneously thrown from the edge of a cliff a distance h0 above the ground. Rock A is thrown vertically upward with speed v0 and rock B is thrown vertically downward with speed v0. Rock A and rock B strike the ground at times tA and tB, respectively. Consider the positive vertical direction to be upward.
(a) On the axes given below, sketch and label graphs of the velocity as a function of time for rock A and rock B. Label the time tB. Times tA and 1/2tA are given on the graph.
The Correct Answer and Explanation is:
To solve this problem, we analyze the motion of two identical rocks thrown from the same height but with different initial velocities. One is thrown upward (rock A) and the other downward (rock B), both with the same speed v0v_0. We’re asked to sketch their velocity vs. time graphs and explain the motion.
🧠 Understanding the Motion:
Rock A (Thrown Upward):
- Initial velocity: +v0+v_0 (positive because it’s upward).
- Due to gravity, the rock slows down at a rate of gg (approximately 9.8 m/s29.8 \, \text{m/s}^2) until it reaches its highest point.
- At the top, its velocity is 0.
- Then it starts to fall downward, accelerating at gg, so its velocity becomes negative.
- It continues accelerating until it hits the ground at time tAt_A.
Rock B (Thrown Downward):
- Initial velocity: −v0-v_0 (negative because it’s downward).
- It accelerates in the same direction as its motion, so its speed increases.
- It reaches the ground at an earlier time tBt_B, where tB<tAt_B < t_A, because it didn’t first travel upward.
📈 Velocity vs. Time Graph:
- The vertical axis is velocity (m/s), with positive values above and negative below.
- The horizontal axis is time, starting from 0.
- For both rocks, the velocity changes linearly with time due to constant acceleration by gravity.
Rock A:
- Starts at +v0+v_0, decreases linearly.
- Crosses velocity = 0 (top of its flight) at time tA2\frac{t_A}{2}.
- Continues to become more negative until time tAt_A.
Rock B:
- Starts at −v0-v_0, becomes more negative linearly.
- Hits the ground at tBt_B, which is before tAt_A.
📍 On the graph:
- Rock A’s line starts at +v0+v_0, slopes down through 0 at tA2\frac{t_A}{2}, ends at a large negative value at tAt_A.
- Rock B’s line starts at −v0-v_0, slopes downward (more negative), and ends at tBt_B, which is between 0 and tAt_A.
✅ Conclusion:
This graph illustrates that gravity affects both rocks equally, but rock A takes longer due to its upward motion first. Both lines are straight (constant acceleration), and the key difference is the starting velocity and total time before hitting the ground.