SOLUTION: A steel ball falls from rest through a height of 2.10 meters . An electronic timer records a time of 0.67 seconds for the fall. c) Calculate the average acceleration of the ball a

Algebra ->  Test -> SOLUTION: A steel ball falls from rest through a height of 2.10 meters . An electronic timer records a time of 0.67 seconds for the fall. c) Calculate the average acceleration of the ball a      Log On


   



Question 1200488: A steel ball falls from rest through a height of 2.10 meters . An electronic timer records a time of 0.67 seconds for the fall.
c) Calculate the average acceleration of the ball as it falls.
d) Suggest reasons why the answer is not exactly 9.81 ms^-2

Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
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A steel ball falls from rest through a height of 2.10 meters .
An electronic timer records a time of 0.67 seconds for the fall.
c) Calculate the average acceleration of the ball as it falls.
d) Suggest reasons why the answer is not exactly 9.81 ms^-2
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Use the standard formula "the height - the time" for the uniformly accelerated move

    h = %28at%5E2%29%2F2,


where "a" is the average acceleration.


In your case it is

    2.10 = %28a%2A0.67%5E2%29%2F2.


It gives

    a = %282%2A2.10%29%2F0.67%5E2 = 9.36 m/s^2.


It is the answer to question (c).


The normal/ordinary value for the gravity acceleration at the Earth surface is 9.81 m/s^2.


The reason for the difference usually is due to the air resistance.

Solved.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
d) Suggest reasons why the answer is not exactly 9.81 ms^-2
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The "constant" varies with location.
Some gauges and software we make (where I work) uses the value at the site.