Phy 201
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: ->->->->->->->->->->->-> : Based off this information given, I conclude that the acceleration is 9.77 m/s^2. I get this by analyzing the problem and determining that the initial velocity is 0m/s, the ‘dt = 0.64 sec, and the ‘ds = 2 meters. With the formula vAve = ‘ds/’dt I get an average velocity of 3.125m/s. And if I multiply the vAve by 2 I can also get the vf = (3.125 *2) = 6.25m/s. From there I can determine the ‘dv = 6.25m/s – 0m/s = 6.25ms, and the ‘dt = 0.64sec, so the aAve = 6.25m/s / 0.64 sec = 9.77m/s/s.
• 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 it is not. If the ball is dropped at v0 = 0m/s, ‘ds = 5meters, and ‘dt = 1.05 meters, the acceleration would then be about 9.07m/s^2. By doing the same exact procedure as above, I get a vAve = 4.76m/s and vf = 9.52. So a = 9.52m/s / 1.05sec = 9.07m/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 observation is consistent with the accepted value of acceleration of gravity of 9.8 m/s^2.
The judgement of consistency in fact depends on the accuracy of the instruments. With very accurate instruments, both values might be considered inconsistent. With instruments of marginal accuracy, it's possible that both would be considered consistent.
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About 15 minutes
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&#Very good responses. Let me know if you have questions. &#