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PHY 121
Your 'cq_1_07.1' 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 falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.
Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> n:
v0 = 0m/sec
`dt = .64sec
`ds = 2m
`ds = (v0 + vf)/2 *`dt
`ds/`dt = (v0 +vf)/2
2(`ds/`dt) = v0 + vf
Vf = 2(`ds/`dt) - v0
Vf = 2(2m/.64sec) - 0m/s
Vf= 2(3.125m/sec)
Vf = 6.25m/sec
aAve = (6.25m/sec-0m/sec)/.64sec
aAve = 9.77m/s^2
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Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?
answer/question/discussion: ->->->->->->->->->->->-> n:
Vf = 2(`ds/`dt) - v0
Vf = 2(5m/1.05sec)
Vf = 9.52m/sec
aAve = (9.52m/s-0m/s)/1.05s
=9.07m/s^2
aAve is not consistent with the first calculation
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Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> n:
The first answer is consistent and only differs because of rounding errors. The second answer is not consistent with the accepted value.
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You calculated the accelerations correctly. Very good.
There's nothing wrong with your answer for consistency either, but here's a little more on that idea:
Consistency depends on the accuracy and precision of the instruments used to make the observations. With a hand-held timer, for example we could easily have a 10% error in timing. If you got 9.07 m/s after timing the fall with a hand-held timer, your result would be consistent with the accepted value.
Using electronic means to obtain the results, on the other hand, it could be that neither observation is consistent.
Since the means of observation aren't specified, of course. So as long as you don't say that the second result is consistent while the first is not, just about any answer is OK.
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