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Phy 201
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** CQ_1_07.1_labelMessages **
7.1 Seed
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: ->->->->->->->->->->->-> :
To find acceleration we have to find the average velocity, use the average velocity to find the final velocity, and then we can find the acceleration:
vAve=2m/0.4s
=3.125m/s
(vf-vi)/2=3.125m/s
vf=6.25m/s
d=vi*t + .5*a*t^2
2m= 0 + .5*a*0.64^2
a=9.77mm/s/s
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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: ->->->->->->->->->->->-> :
We can check this by using the same equation used above to find acceleration, but solve for time:
d=vi*t + .5*a*t^2
5m=0 * 0.5*9.77m/s/s*t^2
t=1.01s
It’s not exact, but it’s within 0.04 cm/s/s
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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: ->->->->->->->->->->->-> :
Yes, the acceleration found in the first problem was 9.77m/s^2 and when rounded to one decimal place it is 9.8m/s^2, the gravity constant.
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&#Very good responses. Let me know if you have questions. &#