course PHY 201
Seed question 7.1A 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
Vavg = 2/.64 = 3.125m/s
Vf = 2*vavg – vo = 2* 3.125 – 0 = 6.25m/s
‘dv = vf –v0 = 6.25m/s – 0 = 6.25m/s
A = ‘dv/’dt = 6.25/.64 = 9.8 m/s/s
• 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:
v0 = 0m/s
‘ds = 5m
‘dt = 1.05s
Vavg = 5/1.05 = 4.76m/s
Vf = 2*4.76 = 9.52m/s
‘dv = 9.52 – 0 = 9.52m/s
A = 9.52 / 1.05 = 9.1m/s/s
This observation is close but not consistent with the prior calculation of acceleration.
Whether or not it is consistent depends on the accuracy of the observations. If the observation is made by a person with a stopwatch, where the error might be +-.1 second, these observations would be consistent. If the observation is made by an accurate and well-synchronized electronic timing device, the results would not be consistent.
• 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 problem is consistent with the accepted value of the acceleration of gravity. The second problem is not consistent with the accepted value of the acceleration of gravity.
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Good, but the question of consistency depends a lot on the accuracy of the instruments used to obtain the measurement. This isn't specified in the problem, so some interpretation is necessary.
See my notes, which you will understand very well.