phy 201
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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: ->->->->->->->->->->->-> : vAve= 2/.64= 3.13m/s
v0 is at rest so = 0m/s
The average velocity= vo+vf/2
We know the average velocity ad the vo so we should be able to find the vf.
3.13m/s= 0m/s+vf/2
=2(3.13)= 6.26 m/s= vf
Now we need to the ‘dv to find acceleration because we already have the ‘dt (.64 s)
‘dv= vf-vo
=6.26-0= 6.26m/s
Acceleration= 6.26/.64= 9.78m/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: ->->->->->->->->->->->-> : 5m/1.05s= 4.76m/s= vAve; I’m not sure if it is consistent or not. I do not understand what I am being asked.
• Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> : Yes, my answer concluded that the acceleration was 9.78m/s^2
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15minutes
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Good start. It won't be hard for you to answer the last two questions.
&#Please compare your solutions with the expanded discussion at the link
Solution
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