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PHY 241
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: ->->->->->->->->->->->-> :
Direct reasoning after one second `dv= a*`dt=-10*1=-10 (Acceleration is down therefore negative.) vf=v0+`dv= 25m/s+ -10m/s=15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
Direct reasoning after 2 seconds `dv= a*`dt=-10*2=-20 vf=v0+`dv= 25m/s+ --20m/s=5m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve=(v0 + vf) / 2=(25 + 5) / 2=15m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds= vAve * `dt=15m/s*2s=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: ->->->->->->->->->->->-> :
It is changing by -10 every second so follow the pattern to see -5m/s at 3 seconds and -15m/s at 4 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: ->->->->->->->->->->->-> :
looking for the `dt at which vf=0
0m/s = 25m/s + -10m/s^2 * `dt
-25m/s=-10m/s^2 * `dt
`dt=2.5s
To find height vAve= 25m/s/2= 12.5m/s
`ds= vAve*`dt=12.5m/s*2.5s=31.25m
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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: ->->->->->->->->->->->-> :
vAve =(v0 + vf) / 2=(25m/s + -15m/s) / 2=5m/s
`ds = vAve * `dt = 5m/s*4= 20m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
At the end of six seconds the velocity will be -35m/s
vAve =(v0 + vf) / 2=(25m/s + -35m/s) / 2=-5m/s
ds = vAve * `dt = -5m/s*4= -20m
If this ball was thrown from ground level then it will be at zero since it cant go through the ground. If thrown from a cliff it will be 20 m below the thrower.
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Very good.
I'm also glad to see you invoking direct reasoning where it is appropriate.
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