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PHY 201
Your 'cq_1_08.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.1_labelMessages **
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
•What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball starts out at 25 m/s and changes velocity by -10 m/s every second.
15m/s
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•What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
average velocity is 20 m/s and the displacement is `ds = vAve `dt = 20 m/s * 1 s = 20 m
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•During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
average velocity for the first 2 seconds is the average 15 m/s of its initial 25 m/s velocity and its final 5 m/s velocity on this interval
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•How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
ds = vAve * `dt = 15 m/s * 2 = 30 m
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•What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
5 m/s * 4 s = 20 m.
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•At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
2.5 s
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•What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> :
12.5 m/s * 2.5 s = 31.25 meters
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•How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
-5 m/s * 6 s = -30 m
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