phy 121
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When you copy a cq problem, just copy the problem. Don't copy the entire form or the instructions. I've been deleting those lines manually, and don't mind doing so on occasion (it only takes a few seconds). However that information includes some lines that could, if I was to miss them (as I would be bound to do eventually), interfere with the proper posting of your work. At the very least they would clutter your work with unnecessary information.
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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:
Acceleration = change in velocity/change in time = 2 meters/.64 sec. / .64 sec. = 1m/3.125sec / .64 sec = 1m/2 sec^2
I see your intent. 2 m / (.64 s) is the average velocity. You divide this by .64 sec to get the acceleration.
However you need to base acceleration on final velocity. The ball starts from rest, so the final velocity will be double the average velocity.
The 3.125 sec isn't correct and isn't in the right place. 2 m / (.64 s) = 3.125 m/s.
It should be easy to correct this, since you're following a valid train of thought. #$&*
Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?
answer/question/discussion:
It would make more sense if a ball dropped took 1.6 sec. from 5 meters based on the previous question.
#$&*
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion:
No, in the first example 2 meters requires over half a second, and in the second example 5 meters are requiring over a second.
#$&*
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