cq_1_071

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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: ->->->->->->->->->->->-> :

The known quantities are: v0=0 m/s; `ds = 2 m; `dt = .64 s

To find the average velocity, you have to solve `ds/`dt = 2 m/.64 s = 3.125 m/s

To find the final velocity, you have to solve (v0 + vf)/2 = vAve

(0 m/s + vf)/2 = 3.125 m/s; vf = 6.25 m/

To find the acceleration, you have to solve `dv/`dt = (6.25 m/s-0 m/s)/.64 s = 9.76 m/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: ->->->->->->->->->->->-> :

The known quantities are: v0=0 m/s; `ds = 5 m; `dt = 1.05 s

To find the average velocity, you have to solve `ds/`dt = 5 m/1.05 s = 4.76 m/s

To find the final velocity, you have to solve (v0 + vf)/2=vAve

(0 m/s +vf)/2 = 4.76 m/s; vf = 9.52 m/s

To find the acceleration, you have to solve `dv/`dt = (9.52 m/s - 0 m/s)/1.05 s = 9.07 m/s^2

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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: ->->->->->->->->->->->-> :

The first one would round to 9.8 m/s^2, so it does work. The second one seems a little low.

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&#Good work. Let me know if you have questions. &#