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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.
** 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: ->->->->->->->->->->->-> :
Given: ‘ds=2m, ‘dt=.64s, v0=0
‘ds = v0*‘dt + .5*a*’dt^2, solve for a
‘ds - v0*’dt = .5 * a * ‘dt^2
(ds - (v0*’dt))/(.5*’dt^2) = a
(2m - (0m/s*.64s))/(.5*(.64s)^2) = 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: ->->->->->->->->->->->-> :
Given: ‘ds=5m, ‘dt=1.05s, v0=0
‘ds = v0*‘dt + .5*a*’dt^2, solve for a
‘ds - v0*’dt = .5 * a * ‘dt^2
(ds - (v0*’dt))/(.5*’dt^2) = a
(5m - (0m/s*1.05s))/(.5*(1.05s)^2) = 9.07m/s^2
This is likely to be consistent with the first observation, given that there could be more error in the 2nd observation.
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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: ->->->->->->->->->->->-> :
Yes, these observations are consistent with the accepted value of gravity at 9.8m/s^2. The only issue is that I have to assume that observation 2 had more error in it.
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30 min
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&#Very good responses. Let me know if you have questions. &#