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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: ->->->->->->->->->->->-> :
I know:
v0 = 0m/s
change in time = 0.64s
displacement = 2m
I need to find vf and a
`ds = (v0 + vf) / 2 * `dt, (`ds / `dt) * 2 - v0 = vf, (2m/0.64s) * 2 - 0m/s = 6.25m/s
vf = 6.25m/s
vf = v0 + a*`dt, (vf / `dt) - v0 = a, (6.25m/s / 0.64s) - 0m/s = 9.8m/s^2
a = 9.8m/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: ->->->->->->->->->->->-> :
`ds = (v0 + vf) / 2 * `dt, (`ds / `dt) * 2 - v0 = vf, (5m / 1.05s) * 2 - v0 = 9.52m/s
vf = 9.52m/s
vf = v0 + a*`dt, (vf / `dt) - v0 = a, (9.52m/s / 1.05s) - 0m/s = 9.07m/s^2
a = 9.1m/s^2
The two values are close. I would say yes.
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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: ->->->->->->->->->->->-> :
I would argue that the first set of data are consistent with the accepted value of the acceleration of gravity. But, the displacement in the second observation changed by 3m (5m - 2m = 3m), which is 1.5 times that of the first value. I believe that since we're talking about uniform acceleration, then the time should have also change by a value of 1.5 time, or 0.64s * 1.5 = 0.96s. If I round the values to 5m/s * 2 = 10m/s, then 10m/s / 1s = 10m/s^2, I'm hovering around the desired 9.8m/s^2. So I would say yes.
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&#Very good responses. Let me know if you have questions. &#