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Phy 231
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: ->->->->->->->->->->->-> :
vo = 0
vave = 2 m / 0.64 sec = 3.125 m / sec
vave = (vf - vo) / 2
vf = 2 * vave = 2 * 3.125 m /sec = 6.25 m /sec
'dv = vf - vo = 6.25 m /sec
a = 'dv /'dt = 6.25 m/sec / 0.64 sec = 9.8 m/sec
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When you divide m/s by s you don't get m/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: ->->->->->->->->->->->-> :
'ds = 5m, 'dt = 1.05 s
vAve = 'ds / 'dt = 4.7619 m/s
vf = 2*vAve = 9.5238 m/s
'dv = vf - vo = 9.5238
a = 'dv / 'dt = 9.07 m/s^2
Awnser is no, the acceleration is not the same.
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Consistency depends on the accuracy of the instruments used. This result could be completely consistent with the preceding if the instruments caused an experimental uncertainty of, say, 15%.
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&#Good responses. See my notes and let me know if you have questions. &#