PHY 201
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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Seed Question 8.1
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:
Since acceleration of gravity is 10 m/s/s downward
then after 1 second the ball should have a velocity of
15 m/s
What will be its velocity at the end of two seconds?
answer/question/discussion:
5 m/s
During the first two seconds, what therefore is its
average velocity?
answer/question/discussion:
Well v0 = 25 and vF = 5
30/2 = 15 m/s is the vAve
How far does it therefore rise in the first two
seconds?
answer/question/discussion:
5-25 = -20
It decelerates 20 m/s
This doesn't tell you how far the ball rises.
What will be its velocity at the end of a additional
second, and at the end of one more additional second?
answer/question/discussion:
In another second I assume it would be traveling back
down at 5 m/s.
I assume that in half a second the ball reached the
top of its path and for a very short amt of time had a
velocity of 0. Then in the other half of a second it
was coming back down toward the earth.
In another second I assume it would be traveling 15
m/s
At what instant does the ball reach its maximum
height, and how high has it risen by that instant?
answer/question/discussion:
At 2.5 seconds the ball has reached its maximum
height.
To find the distance traveled we need to know the
following:
vAve = ?
'dt = 2.5 seconds
vF + v0 = 25/2 = 12.5
12.5 * 2.5 = 31.25 meters was the distance traveled.
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:
After 4 seconds it will be coming back toward the
earth at 15 m/s.
vAve for the first 2.5 seconds is 12.5 m/s
For the next 1.5 seconds it is 7.5 m/s.
I do not know how to find the average velocity for
this.
I do know however that at 1.5 seconds after the ball
is coming back down it has an vAve of 7.5
So 7.5 * 1.5 = 11.25. So subtract this from the
highest point of the ball (31.25) and you get a
heighth of 20 meters after 4 seconds.
So therefore the total distance traveled by the ball
after 4 seconds is 31.25 + 11.25 = 42.5 meters
So then we take 42.5/4 = 10.625 m/s
This is the average velocity after 4 seconds
How high will it be at the end of the sixth second?
answer/question/discussion:
At 5 seconds the ball will be back at its original
point where it was thrown and be traveling at 25 m/s.
So after another second it will be traveling 35 m/s.
The vAve between second 5 and 6 is 30 m/s with 'dt = 1
So it travels 30 meters below the point where it was
thrown. This can be called -30 meters.
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30 mins
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Submission of a revision is optional (no need unless you have questions or comments; I think you'll understand it fine), but check out the link below to see how this problem is set up by first choosing a positive direction, then using + and - signs as appropriate to all quantities.
&#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. &#