Phy 231
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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:
Answer: The ave vel = ds/dt = 2m/.64s = 3.1m/s, acc = 3.1m/s / .64s = 4.8m/s/s.
Acceleration is not equal to the average velocity divided by the time interval. The average velocity has no direct relationship with the acceleration. Please revise this solution in terms of the definition of acceleration.
• 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:
Answer: ave vel = ds/dt = 5m/1.05s = 4.8m/s, acc = 4.8m/s / 1.05s = 4.6m/s/s, which is virtually the same as previous problem, I probably just screwed up the sig figs.
• Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion:
Answer: No but I’m guessing that it depends on how far you fall is what determines the acceleration. For example, I bet that if you fall out of an airplane, you will be accelerating for a period of time and then you will stop accelerating. Your velocity will still be high, but you won’t be speeding up. This is just a guess.
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10 minutes
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You did not calculate your accelerations correctly.
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