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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.
** 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: ->->->->->->->->->->->-> :
( 15 m/s + 0 m/s) / 2 = 7.5m/second * 1.5 second= 11.25 m/second
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7.5 m/s * 1.5 s = 11.25 m, not 11.25 m/s.
This is ave velocity * time interval, which is a correct way to answer the question of how high the object rises.
11.25 m/s is not an answer to the question of 'how high'. 11.25 m is and answer to this question.
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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: ->->->->->->->->->->->-> :
aAve = ( vf - v0) / `dt
vf^2 - 225 m^2/s^2 = 240 m^2/s^2
vf^2 = 465 m^2 / s^2
sqrt(vf^2) = sqrt(465 m^2 / s^2)
vf = 21.56 m/second
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vf = +- 21.56 m/s.
In this situation only the -21.56 m/s result is valid. When the ball first contacts the ground its velocity is downward, which by making acceleration - 10 m/s^2 (a good choice) you have chosen as the negative direction.
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when ball hits the ground
-10 m/s/s = (21.56 m/s - 15 m/s)
`dt = (21.56 m/s - 15 m/s)
`dt = .656 s
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Your equation would be
-10 m/s/s = (21.56 m/s - 14 m/s) / `dt.
The solution to your equation would be -.656 s.
The ball doesn't strike the ground before it is released, so this isn't a correct solution.
The problem is that vf is -21.56 m/s, not 21.56 m/s. If you use -21.56 m/s for vf, you'll get the right result.
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Time for the ball to fall from ground (12 ft)
Time for ball to go above the 12 ft spot and fall back to the 12 again
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Your interval, as you've set this problem up (and you have set it up in the best way), is from release to ground, not from the 12 m mark to the 12 m mark.
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• I must also consider the time taken for the ball to rise above the 12 ft and for it to fall back to the 12 ft mark.
• # - 10 m/s/s = ( 0 m/s - 15 m/s )
-10 m/s/s = -15 m/s
`dt = 1.5 s
- 10 m/s/s = ( 15 m/s - 0 m/s ) / `dt
• `dt * -10 m/s/s = 15 m/s
• `dt = 1.5 s
• 1.5 s + 1.5 s + .656 s = 3.656 s
(the 1.5 + 1.5 added to the 12 equals the 15 meters/second from the problem)
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It's not necessary to do this. Your setup includes the entire interval from release to the ground. You just have to tweak a couple of details to get a really good solution.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
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The ball also has a speed of 5 m/s when its velocity is -5 m/s.
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-10 m/s/s = (5 m/s - 15 m/s) / `dt
= -10 m/s (divide by -10)
`dt = 1 s
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• At what clock time(s) will the ball be 20 meters above the ground?
20 meters - 12 meters = 8 meters
vf^2 = v0^2 + 2 a `ds
vf^2 = 15 m/s^2 + 2 * -10 m/s/s * 8 m
vf^2 = 225 m^2/s^2 -160 m^2/s^2
vf = 8.06 m/s
vAve = `ds / `dt
(15 m/s + 8.06 m/s) / 2 = 8 m / `dt
= 11.53 m/s `dt = 8m
`dt = .69 s
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Very good.
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Ave = (vf - v0) / `dt
-10 m/s/s = ( vf - 15 m/s)
-60 m/s - 15 m/s
-45 m/s = vf
vAve = `ds / `dt
(-45 m/s + 15 m/s) / 2
-15 m/s * 6 s
= -90 m
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*#&!
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You're doing some excellent work here.
You do need to tweak parts of your solution, but overall it's set up very, very well.
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Be sure to include the entire document, including my notes. &#
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