Your 'cq_1_7.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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?
'ds=2m
'dt=.64sec
Vo=O
`ds = v0 `dt + .5 a `dt^2
2=0 x .64 +.5 x a x (.64)^2
a=1.8cm/s^2
Your equation is good but the solution is not 1.8 m/s^2. The solution is closer to 10 m/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?
d=.5at^2
5=.5(1.8)t^2
t=2.02 seconds.
No its not.
You didn't get the correct acceleration in the first question. This check it good; it won't give you identical time of fall but it won't differ by as much as indicated by your result.
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
noI am not getting answers that are consistent with this known value.
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20 min
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&#Good work. See my notes and let me know if you have questions. &#