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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?
9.8m/s^2
a = `dv/`dt = (vf - v0)/`dt = [(2vAve - v0) - v0]/`dt = [(2`ds/`dt - v0) - v0]/`dt = [(2{2m/0.64s} - 0) - 0]/0.64s = 9.8m/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?
No, the acceleration in this case would be 9.1m/s^2.
a = [(2`ds/`dt - v0) - v0]/`dt = [(2{5m/1.05s} - 0) - 0]/1.05s = 9.1m/s^2
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
These observations are only somewhat consistent with the accepted value of the acceleration of gravity. The first acceleration is, but the second one is not.
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20 min.
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Please respond relative to goal of C.
&#Your work looks very good. Let me know if you have any questions. &#