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PHY 121
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.
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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: ->->->->->->->->->->->-> :
v0 = 15m/sec
A = -10m/sec^2
vf = 10m/sec
vAve = (15m/sec+10m/sec)/2 = 12.5m/sec
vf^2 = v0^2 + 2(a*`ds)
(10m/sec)^2 = (15m/sec)^2 + 2(-10m/sec^2 * `ds)
100m^2/sec^2 = 225m^2/sec^2 + 2(-10m/sec^2 *`ds
-125m^2/sec^2 = 2(-10m/sec^2 * `ds)
-62.5m^2/sec^2 = -10m/sec^2 * `ds
`ds = 6.25m
`dt = `ds/vAve
`dt = 6.25m/12.5m/sec
`dt =.5sec
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This isn't consistent with the result of direct reasoning (one reason why you should generally use both and check for consistency).
It should be fairly obvious that at 10 m/s^2 it takes 1.5 sec to come to rest. Since .5 sec isn't consistent with this result, these results need to be reconciled.
The inconsistency is that the ball doesn't reach its highest point until it comes to rest.
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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: ->->->->->->->->->->->-> :
v0 = 10m/sec
`ds = 12m+6.25m = -18.25m
A = 10m/sec^2
`dt1 = .5sec
Vf^2 = v0^2 + 2(a*`ds)
Vf^2 = (10m/sec)^2 + 2(-10m/sec^2 * -18.25m)
Vf^2 = 100m^2/sec^2 + 365m^2/sec^2
Vf^2 = 465m^2/sec^2
Vf = 19.1m/sec
vAve = (19.1m/sec+10m/sec)/2 = 14.55m/sec
-18.25m = 14.55m/sec * `dt
`dt = 1.25sec +(.5sec) = 1.75sec
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = 10m/sec
vf = 5m/sec
a = -10m/sec^2
vf = v0 + a*`dt
5m/sec = 15m/sec + (-10m/sec* `dt)
-10m/sec = -10m/sec*`dt
`dt = 1sec
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At what clock time(s) will the ball be 20 meters above the ground?
`ds = 8m
V0 = 15m/sec
A = -10m/sec^2
Vf^2 = v0^2 + 2(a*`ds)
Vf^2 = (15m/sec)^2 + 2(-10m/sec^2 * 8m)
Vf^2 = 225m^2/sec^2 + -160m^2/sec^2
Vf^2 = 56m^2/sec^2
Vf = 8.06m/sec
vAve = (8.06m/sec+15m/sec)/2 = 11.53m/sec
`dt = `ds/vAve
`dt = 8m/11.53m/sec = .69sec
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
`dt = 6sec
A = -10m/sec^2
V0 = 15m/sec
Vf = 15m/sec + (-10m/sec^2 * 6sec)
Vf = -45m/sec
`ds = (15m/sec+ -45m/sec)/2 * 6sec
`ds =-90m
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I have made several attempts at this problem and am still not certain if I have done it correctly. I am fairly certain that at the beginning of the time interval, v0 should be 15m/sec and that vf should be 10m/sec. I chose to have a equaling a negative 10m/sec since the ball travels in to directions for the problem. The 12m above the ground was throwing me off at first, but I think for the fist problem that this does not factor into the calculation and can be considered a position 0. If all of this is true, I think for the second problem that the 12m does factor in and should be added to the `ds value from the first since the ball would have to travel to it's highest point before it starts to come down. I believe I would also have to add in the `dt value from problem 1 to factor in the time it took to rise to it's final height.
#$&* self-critique
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You're doing a lot of good reasoning, but you've got at least one incorrect assumption. You won't have any trouble fixing it. However be sure to check the link for a couple of alternative ways to set up the solution.
&#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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