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Phy 201
Your 'cq_1_21.1' 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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Copy the problem below into a text editor or word processor.
• This form accepts only text so a text editor such as Notepad is fine.
• You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.
• If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.
• As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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A ball is tossed vertically upward and caught at the position from which it was released.
• Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?
answer/question/discussion: ->->->->->->->->->->->-> :
It will be traveling the same speed as when it was released. In vertical motion:
- >> upward, + >> downward
At the highest vertical point, the time to go up equals that of coming back down.
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• What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion: ->->->->->->->->->->->-> :
I feel that nothing is different than previous answer, given aspecto f air resistance. For example, air resistance will act upon both upward and downward motion; therefore, it will still remain the same speed at both points of location.
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25 minutes
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10/28 8 pm EST
Air resistance is in the same direction as gravitational force for one half of the motion, opposite the gravitational force for the other half.
Check the link.
See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem.