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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 **
Seed 8 1
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: ->->->->->->->->->->->-> :
v0= 25 m/s
a= -10m/s^2
dt=1 s
vf = v0 + a * `dt
vf= 25 m/s -10 m/s^2 *1s
vf=15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf= 25m/s -10 m/s^2 *2s
vf=5 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = (v0 + vf) / 2 * `dt
25 m/s+ 5 m/s)/2 *2s
15m/s*2s
=30 m
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This is how high it will rise during the 2 seconds. That is well done, but it's not the answer to this question.
(v0 + vf) / 2 = 15 m/s, which is the average velocity for the 2 seconds. That is the correct answer to this question.
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve=15 m/s
`ds = v0 `dt + .5 a `dt^2
25 m/s *2 sec +.5(-10m/s/s)(4s^2
30m-20m=
It rises 10 m in 2 seconds
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You found the correct answer to this question in your preceding solution.
This result would agree, but the simpler reasoning is that averaging 15 m/s for 2 s results in a 30 m displacement.
Note that your arithmetic on this step isn't quite right. 25 * 2 is 50, not 30, and your corrected result would indeed be 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 *3s
vf=-5 m/s after 3 s
vf= 25 m/s -10 m/s^2 *4s
-15 m/s afte 4 secs
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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 reached maximum height at 2.5 s
vf=25 m/s *-10m/s^2*2.5
0m/s
Vave= 25+0=25/2=12.5 m/s
12.5 m/s*2.5 s=
31.25 m at max height
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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: ->->->->->->->->->->->-> :
-15+25=10/2= 5 m/s
vAve=5 m/s
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf=v0+a*dt
25m/s -10 m/s^2*6 seconds
25-60
Vf= -35m/s
vAve= -35+25=-10/2= -5 m/s
-5 m/s*6s=
Dt= -30 m
Im assuming it would be on the ground, or if it was thrown up in the air and was coming back down but never hit anything on the way down it would be 30 m below where it was originally thrown up in the air
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Very good, despite a couple of easily corrected errors.
Excellent use of the equations.
No need for a revision but check the discussion at
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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