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Phy 121
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: ->->->->->->->->->->->-> :
vf = 25 m/s + -10 m/s^2* 1 s = 25 m/s + -10 m/s = 15 m/s
I am not sure if this is right. I am assuming that since the ball is moving upward, and the acceleration is moving downward that it is negative.
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•What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = 25 m/s + -10 m/s^2 * 2 s = 25 m/s - 20 m/s = 5 m/s
Still not sure if that is right.
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An acceleration of 10 m/s^2 downward means that an object moving freely upward loses 10 m/s every second, and that an object moving downward gains 10 m/s every second.
Your use of the equations was correct, but not necessary to obtain the correct answers to these questions.
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•During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (25 m/s + 5 m/s) / 2 = 15 m/s
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•How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = `dt * vAve = 2 s * 15 m/s = 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: ->->->->->->->->->->->-> :
vf = 25 m/s + -10 m/s^2 * 3 s = 25 m/s - 30 m/s = -5 m/s
vf = 25 m/s + -10 m/s^2 * 4 s = 25 m/s - 40 m/s = -15 m/s
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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: ->->->->->->->->->->->-> :
vf = 25 m/s + -10 m/s^2 * 2.5 s = 25 m/s - 25 m/s = 0 m/s
vAve = (25 m/s + 0 m/s) / 2 = 12.5 m/s
`ds = `dt * vAve = 2.5 s * 12.5 m/s = 31.25 m
At 2.5 seconds at the height of 31.5 meters.
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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: ->->->->->->->->->->->-> :
vAve = (-15 m/s + 25 m/s) / 2 = 10 m/s / 2 = 5 m/s
`ds = 5 m/s * 4 s = 20 m
Therefore it's average velocity for the first four seconds is 5 m/s and the hight at that point is 20 m high.
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•How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = 25 m/s + -10 m/s^2 * 6 s = 25 m/s + -60 m/s = -35 m/s
vAve = (25 m/s + -35 m/s) / 2 = -10 m/s / 2 = -5 m/s
`ds = -5 m/s * 6 s = -30 m
I don't think this will be negative, so I am going to say it is +30 m.
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*#&!
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If the acceleration continues to be 10 m/s downward, the correct conclusion is -30 meters.
This could be the case, for example, if there was a deep well for the object to fall into.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#