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PHY201
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: ->->->->->->->->->->->-> :
v0=25
a=10
'dt=1
'dv=10*1
'dv=10
vf=v0-'dv=25-10=15
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
v0=25
a=10
'dt=2
'dv=10*2=20
vf=25-20=5
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
Between the interval
25+15/2=20
15+5/2=10
between intial and 2 seconds
5+25/2=15
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
rise is 20
for 2 sec at average 15 m/s?
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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
'dv=10*3
'dv=30
vf=25-30= -5
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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: ->->->->->->->->->->->-> :
25/10=2.5 seconds
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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: ->->->->->->->->->->->-> :
'dv=10*4= 40
25-40=-15m/s or 15 m/s downward
25-15= 10 meters high
ave vel * `dt isn't 10 meters
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
'dv=10*6=60
25-60=-35m/s
25-35=-10 meters or 10 meter below where it started
maybe confusing velocity with position?
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20 minutes
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Solution
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