#$&*
phy201
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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: ->->->->->->->->->->->-> :
a='dv/'dt
'dv=.64s
'dt=2m
a=.64s/2m=
.32m/s
@& .64 seconds is a time interval, not a change in velocity, generally denoted `d v.
2 m is a change in position, i.e., a displacement, denoted `ds.
You do have the right definition of average acceleration, but you need to identify your quantities correctly and reason accordingly.*@
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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: ->->->->->->->->->->->-> :
If we use a propotion we can figure out if the 2 problems are similar
meters/sec=2/.64=5/1.05
2*.64=1.28
.64*5=3.2
The amounts are not equvalent therefore the observations are not consistent with one another
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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: ->->->->->->->->->->->-> :
Neither of the values are consistent with the acceleration of gravity
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*#&!
@& You haven't identified your quantities correctly. You are using the correct definition of aveage acceleration, but you need to use your information differently in order to arrive at a valid solution
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