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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.
** CQ_1_08.1_labelMessages **
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: ->->->->->->->->->->->-> scussion:
25m/s-10m/s(amount velocity changes in one second)= 15m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
15m/s-10m/s(amount velocity changes in one second)= 5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
initial velocity during that 2-second period was 25m/s. Final velocity after 2 seconds was 5m/s.
So vAve=(25m/s+5m/s)2= 15m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
vAve=`ds/`dt
15m/s=`ds/2s
15m/s*2s=`ds
It rises 30m
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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: ->->->->->->->->->->->-> scussion:
At the end of the third second, velocity will be 5m/s-10m/s, or -5m/s, which means it will have reached maximum and begun to fall, reaching 5m/s back downward.
At the end of the fourth second, velocity will be -5m/s-10m/s, or -15m/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: ->->->->->->->->->->->-> scussion:
From the end of the second second to the end of the third second, the ball goes from 5m/s to -5m/s. If acceleration is uniform, it will be at 0m/s (at rest)
halfway between those times.
So, it reaches an instant of rest at maximum height at 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: ->->->->->->->->->->->-> scussion:
initial velocity is 25m/s, final velocity is -15m/s, as we've found
so vAve= (25m/s-15m/s)/2 or 10m/s/2, or 5m/s.
`ds during that interval is vAve*`dt, or 5m/s*4s, or 20m, so it is 20m above its starting point.
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
At the end of the sixth second, velocity will be -35m/s (calculated by taking velocity after 4 seconds and subtracting another 10m/s for each of two additional
seconds). So vAve=(25m/s-35m/s)/2 or -10m/s/2 or -5m/s.
`ds=(-5m/s)(60)=-30m. It would be 30 meters below where it started, if possible. Presuming it was thrown from around ground level, though, it will have
already landed prior to this point.
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10 minutes
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