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PHY 201
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: ->->->->->->->->->->->-> :
vf = v0 + a('dt),
vf = 25 m/s + (-10 m/s^2)(1 s),
vf = 15 m/s.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a('dt),
vf = 25 m/s + (-10 m/s^2)(2 s),
vf = 5 m/s.
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During the first two seconds, what therefore is its average
velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (vf + v0) / 2,
vAve = (25 m/s + 5 m/s) / 2s,
vAve = 15 m/s.
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds = (vf + v0) / 2 * 'dt,
'ds = (25 m/s + 5 m/s) / 2 * 2 s,
'ds = 30 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: ->->->->->->->->->->->-> :
vf = v0 + a('dt),
vf = 25 m/s + (-10 m/s^2)(3 s),
vf = -5 m/s.
vf = v0 + a('dt),
vf = 25 m/s + (-10 m/s^2)(4 s),
vf = -15 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: ->->->->->->->->->->->-> :
The ball will reach its maximum height when v0 equals 0 m/s.
vf = v0 + a('dt),
(vf - v0) / a = 'dt,
(0 m/s - 25 m/s) / (-10 m/s) = 'dt,
2.5 s = 'dt.
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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: ->->->->->->->->->->->-> :
It takes 2.5 seconds to reach the maximum height.
'ds = (vf + v0)/2 * 'dt
'ds = (0 m/s + 25 m/s)/2 * 2.5s,
'ds = 31.25 m high.
Coming back down:
vf = v0 + a('dt)
vf = 0 + (-10 m/s^2)(4s - 2.5s),
vf = -15 m/s.
'ds = (vf + v0)/2 * 'dt
'ds = (-15 m/s + 0 m/s)/2 * 1.5s
'ds = -11.25 m
so 31.25 m - 11.25 m = 20 m above initial position.
vAve = 'ds / 'dt
vAve = (31.25m + 11.25m) / 4 s = 10.625 m/s (???? I'm not sure
about this?????)
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@& Average your initial and final velocities for this interval, which would give you (25 m/s + (-15 m/s) ) / 2 = 5 m/s.
Averaging 5 m/s for 4 s the height would be 20 m.*@
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
If I have correctly solve for the time above (which I doubt) the
ball should hit the ground, or the height at which it was
launched, in 5 seconds.
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*#&!
@& Assuming the ground is there to hit, yes.
Assuming uniform acceleration for 6 seconds, there has to be a well or something for the object to continue falling into.
No revision is necessary, but see my notes and check out the link:
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&#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. &#
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