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Phy 121
Your 'cq_1_08.1' 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: ->->->->->->->->->->->-> :
‘dt = 1
vf = 25 m/s + 10 m/s^2(1 sec) = 35 m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
‘dt = 2
vf = 25 m/s + 10 m/s^2(2 sec) = 45 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
(35 m/s + 45 m/s)/2
(80 m/s)/2
40 m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
(40 m/s)(2 sec)
80 m
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• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
‘dt = 3
vf = 25 m/s + 10 m/s^2(3 sec) = 55 m/s
‘dt = 4
vf = 25 m/s + 10 m/s^2(4 sec) = 65 m/s
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
at the 4 second interval
(65 m/s)/2 = 32.5 m/s
(32.5 m/s)(4 seconds) = 130 meters
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• 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 4 second interval
(65 m/s)/2 = 32.5 m/s
at the 8 second interval
vf = 25 m/s + 10 m/s^2(8 sec) = 105 m/s
(105 m/s)/2 = 52.5 m/s
(52.5 m/s)(8 sec) = 420 meters
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
at the 6 second interval
vf = 25 m/s + 10 m/s^2(6 sec) = 85 m/s
(85 m/s)/2 = 42.5 m/s
(42.5 m/s)(6 sec) = 255 meters
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Unknown
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Good reasoning but the acceleration is downward while the initial velocity is upward. This has implications for the signs. Check the link. I think you'll understand so revision is optional.
&#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. &#