Phy 201
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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:
v0=0m/s
ds=2m
dt= .64s
ds=(vf+vo/2)*dt
2*ds/dt=(vf+v0/2)
2*ds/dt - vo=vf
(2*2m/.64s)*0m/s=vf
4m/.64s-0m/s=vf
6.25m/s-0m/s=vf
6.25m/s=vf
a=dv/dt
a=(6.25m/s-0m/s)/.64s
a=6.25m/s/.64s
a=9.8m/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?
answer/question/discussion:
2*ds/dt=(vf+vo)*dt
vf=2*ds/dt-vo
vf=2.5m/1.05s-0m/s
vf=10m/1.05s
vf=9.52
a=dv/dt=9.5m/s/1.05s
a=9.07m/s^2
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion:
no. I believe the difference is due to a difference in mass? but i'm not sure. I know that an object falls at 9.8m/s^2 due to gravity but i dont understand why these two problems don't follow that theory.
These are said to be observations, which can be subject to errors and uncertainties. Depending on the accuracy of the experiment, these results might or might not be consistent with the accepted value of g.
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10 min.
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had a little confusion with this assignment but understand what it's concept was.
&#Good work. See my notes and let me know if you have questions. &#