An object is shot vertically from top of a building 60 m above the ground with an initial speed vo=27 m/s (at point O). Define the ground as y=0 and call “UP” to be the positive y-axis. Assume no air. Responses should be based on Graphical analysis. No credit given for analytical solution a. Without using the kinematic equations determine the velocity at t=0 to t=4s. Show work and explain t (5) 0 1 2 3 4 V(m/s) b. Based on the values in part a draw the velocity vs time graph for this motion velocity (m/s) 30 20 10 0 -10 234 5 -20 -30 time (s) c. Based on the table or graph determine how long does it take to reach the maximum height? Show work and explain.
The Correct Answer and Explanation is:
Let’s break down the problem step by step and focus on graphical analysis.
a. Determine the velocity at different times (t = 0, 1, 2, 3, 4 seconds):
Since the object is shot vertically upward from the building at v0=27 m/sv_0 = 27 \, \text{m/s}v0=27m/s, and the motion is under gravity, we know the object will slow down until it stops momentarily (at the maximum height) and then start falling back. However, since we’re using a graphical analysis, we can infer a lot of information just by understanding the motion and gravity’s effects on velocity.
- Initial velocity (at t=0t = 0t=0): Since the object is shot upwards with a velocity of v0=27 m/sv_0 = 27 \, \text{m/s}v0=27m/s, at t=0t = 0t=0, the velocity is 27 m/s27 \, \text{m/s}27m/s. So the velocity at t=0t = 0t=0 is 27 m/s.
- At t=1st = 1st=1s: The object will experience a constant downward acceleration due to gravity, which is approximately 9.8 m/s29.8 \, \text{m/s}^29.8m/s2. So after 1 second, the velocity will decrease by 9.8 m/s9.8 \, \text{m/s}9.8m/s. Therefore, at t=1 st = 1 \, \text{s}t=1s, the velocity is approximately v1=27−9.8=17.2 m/sv_1 = 27 – 9.8 = 17.2 \, \text{m/s}v1=27−9.8=17.2m/s.
- At t=2st = 2st=2s: After another second, the velocity will decrease by another 9.8 m/s9.8 \, \text{m/s}9.8m/s, so at t=2 st = 2 \, \text{s}t=2s, the velocity is v2=17.2−9.8=7.4 m/sv_2 = 17.2 – 9.8 = 7.4 \, \text{m/s}v2=17.2−9.8=7.4m/s.
- At t=3st = 3st=3s: After 3 seconds, the velocity will decrease further by 9.8 m/s9.8 \, \text{m/s}9.8m/s, so at t=3 st = 3 \, \text{s}t=3s, the velocity is v3=7.4−9.8=−2.4 m/sv_3 = 7.4 – 9.8 = -2.4 \, \text{m/s}v3=7.4−9.8=−2.4m/s. This negative velocity indicates that the object is now moving downwards after reaching its highest point.
- At t=4st = 4st=4s: Finally, after 4 seconds, the object will have accelerated downward further, so the velocity is v4=−2.4−9.8=−12.2 m/sv_4 = -2.4 – 9.8 = -12.2 \, \text{m/s}v4=−2.4−9.8=−12.2m/s.
So the table would look like this:
| t (s) | v (m/s) |
|---|---|
| 0 | 27 |
| 1 | 17.2 |
| 2 | 7.4 |
| 3 | -2.4 |
| 4 | -12.2 |
b. Draw the Velocity vs Time graph:
We can plot the velocity values from the table onto a graph. The y-axis represents velocity (in m/s) and the x-axis represents time (in seconds). Here’s how the graph would look:
- At t=0t = 0t=0, the velocity is +27 m/s+27 \, \text{m/s}+27m/s, which starts the curve at the top of the graph.
- At t=1st = 1st=1s, the velocity drops to +17.2 m/s+17.2 \, \text{m/s}+17.2m/s.
- At t=2st = 2st=2s, the velocity further drops to +7.4 m/s+7.4 \, \text{m/s}+7.4m/s.
- At t=3st = 3st=3s, the velocity becomes negative at −2.4 m/s -2.4 \, \text{m/s}−2.4m/s, indicating downward motion.
- At t=4st = 4st=4s, the velocity continues in the downward direction to −12.2 m/s -12.2 \, \text{m/s}−12.2m/s.
The graph will be a straight line with a negative slope (because gravity is acting uniformly) and crossing the time-axis at the point where the object reaches its highest point (maximum height).
c. Determine how long it takes to reach the maximum height:
The maximum height occurs when the velocity becomes zero. Based on the velocity-time graph, the object starts at a velocity of 27 m/s27 \, \text{m/s}27m/s and slows down at a constant rate due to gravity. The velocity reaches zero when the object stops moving upwards and starts falling.
From the table, we see that the velocity decreases linearly with time. The object reaches v=0 m/sv = 0 \, \text{m/s}v=0m/s between t=2 st = 2 \, \text{s}t=2s and t=3 st = 3 \, \text{s}t=3s. Since the object’s velocity is decreasing at a constant rate, we can estimate the time to reach the maximum height by interpolation between the values for t=2st = 2st=2s and t=3st = 3st=3s.
- At t=2 st = 2 \, \text{s}t=2s, the velocity is 7.4 m/s7.4 \, \text{m/s}7.4m/s, and at t=3 st = 3 \, \text{s}t=3s, the velocity is −2.4 m/s-2.4 \, \text{m/s}−2.4m/s.
- Since the velocity decreases linearly, it takes approximately 2.7 seconds2.7 \, \text{seconds}2.7seconds to reach the maximum height, where the velocity is zero.
Thus, the time to reach the maximum height is approximately 2.7 seconds.
