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.
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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: ->->->->->->->->->->->-> : I think it will just have the acceleration of gravity, 9.8 m/s^2. `ds = v0`dt + .5a`dt^2; a = (`ds – v0`dt) / [.5 * (`dt^2)]; a = (2 m – 0 m/s * .64 s) / [.5 * (.64 s)^2]; a = 2 m / .2048 s^2; a = 9.77 m/s^2 or 9.8 m/s^2 the acceleration of gravity.
• 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: ->->->->->->->->->->->-> : I think this should still have an acceleration of 9.8 m / s^2. `ds = v0`dt + .5a`dt^2; 5 m = 0 + .5 * 9.8 m / s^2 * (1.05 s)^2; 5 m = 5.4 m; Approximately
• Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> : Yes
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20 minutes
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I'm sorry I am resubmitting this. I submitted it earlier in a Submit Work Form because I was interrupted and just forgot what I was doing.
No problem at all. You got the first acceleration; you should have calculated the second in the same way. See the discussion at the link; no need at all to submit a revision. You'll understand.
&#Please compare your solutions with the expanded discussion at the link
Solution
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