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Phy 241
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.
** CQ_1_07.1_labelMessages **
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: ->->->->->->->->->->->-> :
v_ave = 'ds/'dt
= 2 m/ .64 s
= 3.125 m/s
v_0 = 0
v_f = 6.25 m/s
'ds= 2 m
'dt = .64 s
vf = v0 + a * `dt
a = (v_f - v_0) / 'dt
= 6.25 m/s / .64 s
= 9.77 m/s^s
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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: ->->->->->->->->->->->-> :
v_ave = 'ds/'dt
= 5 m/ 1.05 s
= 4.76 m/s
v_0 = 0
v_f = 9.52 m/s
'ds = 5m
'dt = 1.05 s
a = (v_f - v_0) / 'dt
= 9.52 m/s / 1.05 s
= 9.07 m/s^2
The percent difference betweent the two solutions is 7%, so yes. For such a small amount of
observed examples, 7% should show some consistency, but any more than that, I would say no.
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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: ->->->->->->->->->->->-> :
OH! absolutely. 9.8 m/s^2 is the accepted acceleration of gravity, so 9.8, 9.1 same thing.
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&#Very good responses. Let me know if you have questions. &#