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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: ->->->->->->->->->->->-> :
The acceleration of the given ball is 9.77m/s^2 because the average velocity of the3.125m/s. Assuming that the initial velocity is 0m/s and the final velocity is double the average velocity which is 6.25 because 3125*2=6.25. The change in velocity is 6.25m/s divided by the change in time which is .64s gives us the acceleration of 9.77m/s^2 because 6.25m/s / .64s = 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: ->->->->->->->->->->->-> :
No, this is not consistent with the observation that a ball dropped from the height of 5 meters will reach the ground in 1.05 seconds because this acceleration is 9.07 rather than 9.77. This is because the average velocity is 4.76m/s because 5m/1.05s = 4.76m/s. Given that the initial velocity is 0m/s and the final velocity is double the average velocity which is 9.52m/s because 4.76m/s * 2 = 9.52m/s. The change in velocity is 9.52m/s divided by the change in time which is 1.05 seconds gives us the acceleration of 9.07m/s^2 because 9.52m/s / 1.05s = 9.07m/s^2.
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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: ->->->->->->->->->->->-> :
The first example is consistent with the accepted value of 9.8m/s^2, but the seconds is not because its acceleration is varies too much from this accepted value.
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&#Very good work. Let me know if you have questions. &#