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course Phy 121
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:
ds=v0*dt+.5a(dt)^2
[ds-(v0)(dt)]/[(.5)(dt^2)]=a
[(2m-(0m/s)(.64s)]/[(.5)(.64s)^2]=a
a=9.77m/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:
vAve=(vf+v0)/s
4.76m/s=(vf+0m/s)/2
9.52m/s=vf
a=(dv)/(dt)
a=(9.52m/s-0m/s)/(1.05)=9.07m/s^2
No it appears that this second calculation does not match the first.
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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 calculation is very close to the accepted value while the second one is not.
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&#This looks very good. Let me know if you have any questions. &#