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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.
** CQ_1_07.1_labelMessages **
Samantha Rogers
PHY 201
Seed 7.1
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 = `ds / `dt
vAve = 2 meters / .64 sec
vAve = 3.125 meters / sec
vAve = (vf + v0) /2
3.125m/s = (vf) /2
vf = 6.25m/s
To find acceleration I use the following equation:
vf = v0 + a `dt
vf - v0 = a `dt
a = (vf - v0) / `dt
a = (6.25m/s - 0m/s) / .64s
a = (6.25m/s) / .64s
a = 9.77 m/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: ->->->->->->->->->->->-> :
vAve = 5m / 1.05s
vAve = 4.76m/s
vAve = (vf + v0) / 2
4.76m/s = (vf + 0) / 2
4.76m/s = vf / 2
vf = 9.52m/s
a = (vf - v0) / `dt
a = (9.52m/s - 0) / 1.05s
a = 9.52m/s / 1.05s
a = 9.067m/s^2
For this my acceleration value was not as close to 9.77m/s^2. However it is still relatively close to 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: ->->->->->->->->->->->-> :
Yes, my answers were relatively close to this accepted value. (The first answer was closer)
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*#&!
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