#$&*
phy 231
Your 'cq_1_07.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.
Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
Using the equation: `ds = v0*`dt + .5a*`dt^2
We can rearrange the values to equal a: (`ds - v0*`dt) / (.5*`dt^2) = a
(2m - 0*.64) / (.5 * .64^2) = 2 / .2048 = 9.77 m/s^2
#$&*
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: ->->->->->->->->->->->-> :
Use the same equation as last time and plug in numbers:
(`ds - v0*`dt) / (.5*`dt^2)= a
(5m - 0*1.05) / (.5 * 1.05^2) = 5 / .55125 = 9.07 m/s^2
(so nearly the same)
#$&*
Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> :
The first values result in a closer value to the acceleration due to gravity; the second is off by several tenths, but is close to being consistent as well.
#$&*
** **
25 min
** **
@& Good.
Whether the two are consistent would depend on the precision of the setup and the timing. For the precision we are usually able to achieve in this course, we would probably say that the two are indeed consistent.*@