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Phy 121
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 **
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: ->->->->->->->->->->->-> :
vf = v0 + a * `dt
vf = 25m/s + (-10m/s^2) * 1s = 25m/s - 10m/s = 15m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = 25m/s + (-10m/s^2) * 2s = 25m/s - 20m/s = 5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
Assuming that the acceleration is linear(not given an initial or final position though I could find them), vAve = (v0 + vf) / 2 = (25m/s + 5m/s) / 2 = 15m/s. Since I wasn't 100% sure it was linear, I solved the first equation for `ds = (v0 + vf) / 2 * `dt = 30m, vAve = `ds / `dt = 30m / 2s = 15m/s, so it presumes to be linear.
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = vAve * `dt = (15m/s) * 2s = 30m rise over 2 seconds at an average velocity of 15m/s.
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Good. It is valid to assume linearity, since we are using the (constant) acceleration of gravity.
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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 = 25m/s + (-10m/s^2) * 3s = 25m/s - 30m/s = -5m/s
vf = 25m/s + (-10m/s^2) * 4s = 25m/s - 40m/s = -15m/s
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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: ->->->->->->->->->->->-> :
The ball will reach it's maximum height at the point where it touches the x axis, meaning that it is traveling at 0m/s(zero acceleration). At that point, it's risen a total of -25m with a run of 2.5s = -10 slope. vf = 25m/s + (-10m/s^2) * 2.5s = 25m/s - 25m/s = 0m/s, so at 2.5s the ball has reached its 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: ->->->->->->->->->->->-> :
vAve = (v0 + vf) / 2 = (25m/s + -15m/s) / 2 = 5m/s at the end of the 4s mark.
`ds = vAve * `dt = 5m/s * 4s = 20m
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (v0 + vf) / 2 = (25m/s + -35m/s) / 2 = -5m/s at the end of the 6s mark.
`ds = vAve * `dt = -5m/s * 6s = -30m
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&#Good work. See my notes and let me know if you have questions. &#