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PHY 231
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: ->->->->->->->->->->->-> :
'ds=25 m/s*1s+.5*-10 m/s^2 *(1s)^2=20m
20m/1s=20 m/s
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@& If the ball's velocity changes at the rate of 10 m/s^2, then it will change by 10 m/s in one second.
The rising ball slows, so the velocity will clearly be 15 m/s.
That noted, there's nothing at all wrong with your equation or your conclusion that the ball travels 20 m during this 1-second interval. However that does not justify your conclusion that after 1 second the ball is moving at 20 m/s.
Don't let me put too much pressure on you, but you're doing excellent work and I believe you can resolve these discrepancies. Give yourself a reasonable time limit and see if you can; if not, then send me your best thinking with some additional questions and we'll go from there.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds=25 m/s*2s+.5*-10 m/s^2 *(2s)^2=30m
30m/2s=15 m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
(20 m/s-15 m/s)/2=17.5 m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
17.5 m/s*2s=35 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: ->->->->->->->->->->->-> :
'ds=25 m/s*3s+.5*-10 m/s^2 *(3s)^2=30m
30m/3s=10 m/s
'ds=25 m/s*4s+.5*-10 m/s^2 *(4s)^2=20 m
20m/4s=5 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: ->->->->->->->->->->->-> :
'ds=v0*^' dt+.5*a*^' dt^2
'dv=0 m/s=v0+a*^' dt
'dt=-v0/a=-((25m/s)/(-10m))/s^2 =2.5s
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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: ->->->->->->->->->->->-> :
(20m+15m+10m+5m)/4s=12.5 m/s
12.5 m/s*4=50 m
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds=25 m/s*6s+.5*-10 m/s^2 *(6s)^2=120 m
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20 min
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@& You're doing a lot of good things here, but you are misinterpreting some of your results.
Check my note on your first result.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@