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.
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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.
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Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
we know that vAve = ds/dt, so vAve = 2 m/.64 s = 3.125 m/s or roughly 3.1 m/s
since v0 = 0, then vf must be
(0 + vf)/2 = 3.125 m/s
vf = 6.25 m/s
since a = dv/dt, then a would be the change in velocity (6.25 - 0) divided by change in time, or .64 sec, so (6.25m/s)/.64 s = 9.77 m/s^2 or roughly 9.8 m/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: ->->->->->->->->->->->-> scussion:
well, using the same format as above, we can begin by checking
vAve = 5 m/1.05 s = 4.76 m/s
and (0 + vf)/2 = 4.76 m/s
vf = 9.52 m/s
and a = (9.52 - 0)/1.05 = 9.07 m/s^2
This does not corroborate our previous results.
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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: ->->->->->->->->->->->-> scussion:
Yes, the result of the first problem yields 9.8m/s^2, but the second one does not.
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12 minutes
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&#Good responses. Let me know if you have questions. &#