phy 201
Your 'cq_1_7.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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: average acceleration = change in velocity/ time elapsed so 2m/.64sec = 3.125m/s
2 m is a displacement, and has units of displacement; it is not a change in velocity.
m/s is a unit of velocity, not a unit of acceleration.
You have divided displacement by elapsed time, which gives you the average velocity, not the 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: it is consistent with the observation which concludes that a ball dropped from 5 meters reaches the ground in 1.05 sec because if you multiply the .64 * 5 = you get 3.2 m/s which is very close to the 3.125 m/sec from above.
• Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: yes at a given location on the earth and in the absence of air resistance, all objects fall with the same constant acceleration. And the acceleration is constant.
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25min
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You haven't correctly calculated your accelerations.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#