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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: ->->->->->->->->->->->-> :
After one second, the ball should have lost 10 m/s, so it should be traveling 15 m/s.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
At the end of 2 seconds, it should have lost another 10 m/s, so it should be traveling 5
m/s.
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
The initial velocity is 25 m/s and the final velocity after 2 seconds is 5 m/s. To find
the average, you add the two numbers and divide by 2. 25 m/s + 5 m/s = 30 m/s. 30 m/s
/ 2 = 15 m/s
The average velocity is 15 m/s for the first 2 seconds.
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
It rises 30 m in the first two seconds.
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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: ->->->->->->->->->->->-> :
After an additional second, it will have reached its peak and will be starting back
down. This gives it a velocity of 5 m/s - 10 m/s = -5 m/s.
At the end of one more second, it will be going -5 m/s - 10 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: ->->->->->->->->->->->-> :
It will reach its maximum height at 2.5 seconds. This is because it changes direction
halfway between 2 and 3 seconds. It will be at 13.25 m at that time. To get that
answer, I averaged the velocity starting at 25 m/s and ending at 0 m/s. This gave me an
average velocity of 12.5 m/s. At 12.5 m/s times 2.5 s, the height is 31.25 m.
However,the picture that I drew, shows the following:
Second 1 - v0 = 25 m/s, vf = 15 m/s, vAve = 20 m/s, for a total of 20 m in that second.
Second 2 - v0 = 15 m/s, vf = 5 m/s, vAve = 15 m/s, for a total of 15 m in that second
Second 3 - v0 = 5 m/s, vf = -5 m/s, vAve doesn't help, but it's going up 5 meters and
then back down 5 meters at this time. This total is 45 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: ->->->->->->->->->->->-> :
The average velocity for the first four seconds is 5 m/s. You take the 25 m/s and add
the -15 m/s, to get 10 m/s and divide that by 2.
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
I think it's done by the end of the sixth second. My sketch shows it's done at 5
seconds.
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*#&!*#&!
&#This looks very good. Let me know if you have any questions. &#