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phy121
Your 'cq_1_08.2' 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_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2 a `ds
(vf^2-v0^2)/2*a=’ds
‘ds= (0-15m/s^2)/ 2*(-10m/s^2)
‘ds= -225m/s / -20m/s^2
‘ds= 11.25m
However, that’s from the point of release, it would be 11.25m+12m= 23.25m above the ground.
`ds = (vf + v0) / 2 * `dt
‘ds/(vf + v0) / 2= ‘dt
‘dt= 11.25m/(0m/s+15m/s)/2
‘dt= 11.25m/7.5m/s
‘dt= 1.5s
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2 a `ds
vf^2= 0m/s^2+ 2*(10m/s^2)*23.25m
vf= sqrt465m/s
vf= 21.6m/s when it hits the ground.
`ds = (vf + v0) / 2 * `dt
‘ds/(vf + v0) / 2= ‘dt
‘dt= 23.25m/(21.6m/s+0m/s)/2
‘dt= 2.15s from the highest point to the ground.
From the toss you would take the time from release to the highest point, plus the time from the highest point to the ground, therefore:
1.5s+2.15s= 3.65s to reach the ground from the initial toss.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Before max height:
vf= v0+a*'dt
(vf-v0)/a= ‘dt
‘dt= (5m/s-15m/s)/(-10m/s^2)
‘dt= -10m/s / -10m/s^2
‘dt= 1s to reach 5m/s after the initial toss and before the max height.
From max height to ground:
vf= v0+a*'dt
(vf-v0)/a= ‘dt
(5m/s-0m/s)/10m/s^2= ‘dt
‘dt= 5m/s / 10m/s^2
‘dt= 0.5s to reach 5m/s after max height has been reached.
To find the time from the initial toss, we have to add that number to the number it took to reach max height.
0.5s+1.5s= 2s
Therefore at 1s and 2s, the ball will be travelling at 5m/s.
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Note that you can also analyze the motion from the first instant to the last as a single interval. For this interval:
Let upward be positive.
Initial velocity is +15 m/s^2, acceleration is -9.8 m/s^2 and displacement is -12 m.
Solving for this interval you should get the same final velocity and time interval as in your original solution.
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Since the thrower is 12m above the ground, we need to find the difference from the 20m and the point where the ball is thrown:
20m-12m= 8m, distance travelled from throw
`ds = v0 `dt + .5 a `dt^2
‘dt= sqrt(v0-‘ds/0.5*a)
‘dt= sqrt((15m/s-8m)/(0.5*10m/s^2))
‘dt= sqrt(1.4s)
‘dt=1.18s to reach 20m from release.
If the max height of the ball was 23.25m, we can subtract 20m from that number to determine how far the ball travelled from max height:
23.25m-20m= 3.25m
`ds = v0 `dt + .5 a `dt^2
‘ds= 0.5*a*’dt^2
3.25m= 0.5*10m/s^2*’dt^2
‘dt^2= 0.65s
‘dt= sqrt0.65s
‘dt= 0.8s from max height.
We have to add that number to the time it took to reach max height to get total clock time:
1.5s+0.8s= 2.3s
Since were trying to find the height of the ball after 6 second, we can subtract the toss to max height time to find how long it travels downward:
6s-1.5s= 4.5s
`ds = v0 `dt + .5 a `dt^2
‘ds= 0+0.5*-10m/s^2*4.5s^2
‘ds= -101.25m
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35 minutes
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sept 26, 4:23pm
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Good work. You are using the equatoins very effectively.
See also the discussion at the link provided below for a reasoning process that is more conceptual in its approach.
Ignore any reference to resubmitting, but do let me know if you have questions.
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&#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. &#