Phy 231
Your 'cq_1_7.1' 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: We know from the problem that v0 = 0 m/s, 'ds = 2 m and that 'dt = 0.64 s. Using the formula: 'ds = v0 * 'dt + .5a * 'dt^2, we can find the acceleration.
'ds = v0 * 'dt + .5a * 'dt^2
2m = 0 m/s * 0.64s + 0.5a * (0.64s)^2
2m = 0.5a * .4096 s^2
2m = a * 0.2048s^2
a = 9.77 m/s^2
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: Using the same formula we get:
'ds = v0 * 'dt + .5a * 'dt^2
5m = 0 m/s * 1.05s + 0.5a * (1.05s)^2
5m = 0.5a * 1.1025s^2
5m = a * 0.55125 s^2
a = 9.07 m/s^2
This answer seems slightly off from what we would expect. The answer is a good bit lower than gravity: 9.8 m/s^2
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion:
The first answer definitely is. The second answer is slightly off - I would expect it to be higher then 9.07 m/s^2. I suppose it's still fairly close though.
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15 minutes
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The question of consistency depends on the accuracy of the measurements. A 5% uncertainty in measuring time interval could conceivably explain the discrepancy in these results.