phy121
Your 'cq_1_8.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion:
v0 + (a * ‘dt) = vf
25 m/s + (-10 m/s^2 * 1 s) = vf
25 m/s + (-10 m/s) = vf
15 m/s = vf
• What will be its velocity at the end of two seconds?
answer/question/discussion:
25 m/s + (-10 m/s^2 * 2s) = vf
5 m/s = vf
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
vAve = (25 m/s – 5 m/s) /2
This is not correct.
vAve = 20 m/s / 2
vAve = 10 m/s
• How far does it therefore rise in the first two seconds?
answer/question/discussion:
vAve * ‘dt = ‘ds
10 m/s * 2 s = ‘ds
20 m = ‘ds
• What will be its velocity at the end of an additional second, and at the end of one more additional second?
answer/question/discussion:
Since I have determined (details in next ?) that the maximum altitude is reached at 2.5 seconds, then the velocity at that point is 0, and acceleration becomes positive for the remaining time as the object begins to descend.
this is only so if you change the positive direction from up to down; the ball still accelerates in the downward direction
0 m/s + (10 m/s^2 * .5s) = vf
0 m/s + 5 m/s = vf
5 m/s = vf
0 m/s + (10 m/s^2 * 1.5s) = vf
0 m/s + 15 m/s = vf
15 m/s = vf
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
25 m/s + (-10 m/s^2 * ‘dt) = 0m/s
-10 m/s^2 * ‘dt = -25 m/s
(-10 m/s^2 * ‘dt) / -10 m/s = -25 m/s / -10 m/s^2
‘dt = 2.5 s
vAve * ‘dt = ‘ds
10 m/s * 2.5 s = ‘ds
good thinking but your average velocity is not correct
25 m = ‘ds
• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion:
At the end of the fourth second, the object will have been traveling down for 1.5 sec at an average velocity of (15 m/s – 0 m/s) /2, or 7.5 m/s. Multiply vAve (7.5 m/s) by ‘dt (1.5 s) to get ‘ds from the maximum height (11.25 m). Now, we add this to the distance traveled to get to the maximum height (25 m), and get a total distance of 36.25 m.
To get average velocity for the entire trip, I would take total distance (36.25 m) and divide by total time (4 s) to get 9.0625 m/s.
• How high will it be at the end of the sixth second?
answer/question/discussion:
The sixth second is 3.5 seconds after the maximum point, so the object has accelerated for 3.5 s from 0 at a rate of 10 m/s^2. We multiply a * ‘dt = ‘dv, or 3.5 s * 10 m/s^2 = 35 m/s. Divide (35 m/s – 0 m/s) by 2 to get vAve. vAve * ‘dt = ‘ds. 17.5 m/s * 3.5 s = 61.25 m from maximum height. If the object started (was thrown) from ground level, it would have reached the ground before 6 seconds, so its height would be 0. From its
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20 min
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You have some incorrect average velocities. Most of your reasoning is correct, but there's more to learn about this situation.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#