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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?
The ball accelerates uniformly from rest so the final velocity is double the
average velocity.
-
vAve = `ds/`dt = 2m/.64s = 3.1m/s
so
-
vf = 2*vAve = 3.1m/s *2 = 6.2m/s
so
-
aAve = `dv/`dt = 6.2m/s / (.64s) = 9.7m/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?
In this case
so
so
-
aAve = 9.6m/s / 1.05 s = 9.1m/s^2
Are these observations consistent with the accepted value of the acceleration
of gravity, which is 9.8 m / s^2?
This depends on how accurate you think the observations are.
The two results for acceleration differ by about 7%.
Assuming that the uncertainty in distance measurement is insignificant,
a 3.5% uncertainty in measuring the time interval will result in a 7%
uncertainty in the calculated acceleration.
This is because in the process of finding acceleration we divide twice
by the time interval, once when finding average velocity and once when
finding average rate of change of velocity. The uncertainty is
therefore compounded. The first division results in a 3.5%
uncertainty; the second adds another 3.5% to the uncertainty.
If the timing uncertainty is less than about 3.5% then the difference in
accelerations is probably significant. If the timing uncertainty is more
than about 3.5% then the difference is probably not significant.
There's more to it than this. However to fully answer the question of
significance would require more use of statistics than is appropriate to this
course, for which a prior statistics course is not a prerequisite.