Question 1200488
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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 = {{{(at^2)/2}}},


where "a" is the average acceleration.


In your case it is

    2.10 = {{{(a*0.67^2)/2}}}.


It gives

    a = {{{(2*2.10)/0.67^2}}} = 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.
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Solved.