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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.
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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: ->->->->->->->->->->->-> :
a= (vf-v0)/’dt
10m/s/s= (vf-25m/s)/1s
Multiply both sides by 1
10m/s/s= vf-25m/s
Add 25m/s to both sides
35m/s=vf
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
We will do the same equation, but plug in 2 seconds instead of 1.
10m/s/s= (vf-25m/s)/2s
20m/s=vf-25m/s
45m/s=vf
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= (vf+v0)/2
vAve= (45+25)/2
vAve=35 m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= ‘ds/’dt
35m/s= ‘ds/ 2s
(multiply both sides by 2)
70m= ‘ds
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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: ->->->->->->->->->->->-> :
10m/s/s= (vf-25m/s)/3s
30m/s=vf-25m/s
55m/s= vf (after an additional second)
10m/s/s= (vf-25m/s)/4s
40m/s=vf-25m/s
65m/s=vf (after one more additional second)
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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: ->->->->->->->->->->->-> :
y will= max height
y= (vf-v0)^2/2a
(0-25m/s)^2/2(10m/s/s)=625/20=31.25 meters, which is the max height the ball reaches.
I can find the ‘dt of this height by:
vAve= ‘ds/’dt
12.5= 31.25/ ‘dt
12.5 ‘dt=31.25
‘dt= 2.5 seconds
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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= ‘ds/ ‘dt
12.5m/s= ‘ds/ 4s
50m= ‘ds
So if it only goes up 31.25 m then it comes back down 18.75m , so it is only 12.5m high.
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
12.5m/s= ‘ds/6s
75m/s= ‘ds
---The ball is already back on the ground.
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You didn't account for the fact that the acceleration of gravity and the initial velocity are in different directions. Check the link; no need for revision unless you see a need for it. I'm sure you'll understand.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your 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. &#