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(52781) (Show Source):
You can put this solution on YOUR website! .
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
~~~~~~~~~~~~~~~~~~~
Use the standard formula "the height - the time" for the uniformly accelerated move
h = ,
where "a" is the average acceleration.
In your case it is
2.10 = .
It gives
a = = 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.
You can put this solution on YOUR website! d) Suggest reasons why the answer is not exactly 9.81 ms^-2
---------------
The "constant" varies with location.
Some gauges and software we make (where I work) uses the value at the site.
