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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.
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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: ->->->->->->->->->->->-> :
For this instance the acceleration will be negative.
vf = v0 + a * ‘dt
vf = 25m/s + -10m/s^2 * 1s
vf = 25m/s -10m/s
vf = 15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a * ‘dt
vf = 25m/s + -10m/s^2 * 2s
vf = 25m/s - 20m/s
vf = 5m/s
Therefore, the final velocity after the given two seconds would be 5m/s.
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (v0 + vf) / 2
vAve = (25m/s + 5m/s) /2
vAve = 30m/s / 2
vAve = 15m/s
Therefore, the average velocity for the first two seconds was 15m/s.
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball rises 30 meters in the first two seconds because the average velocity is 15m/s and traveling at that speed for 2 seconds, so 15m/s * 2s = 30m.
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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 = v0 + a * ‘dt
vf = 25m/s + -10m/s^2 * 3s
vf = 25m/s - 30m/s
vf = -5m/s
The velocity at the end of 3 seconds is -5m/s.
vf = v0 + a * ‘dt
vf = 25m/s + -10m/s^2 * 4s
vf = 25m/s - 40m/s
vf = -15m/s
The velocity at the end of 4 seconds is -15m/s.
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Good but it doesn't take the equations to calculate those velocities. Every second, there is a -10 m/s^2 change in the velocity.
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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: ->->->->->->->->->->->-> :
vf = v0 + a * ‘dt
vf = 25m/s + -10m/s^2 * 2.5s
vf = 25m/s - 25m/s
vf = 0m/s
The ball reaches its maximum height at 2.5s. At this point it will have risen 31.25m because the average velocity for this time period is 12.5m/s and at that rate for 2.5s the ball will travel 31.25m.
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Be sure you also know how to reason this out directly in terms of your understanding of the meanings:
To slow by 25 m/s at -10 m/s/s takes 2.5 seconds for reasons that should be obvious.
During that time the average velocity will be 12.5 m/s.
12.5 m/s * 2.5 s = 31.25 m.
The equations are often necessary, but direct reasoning is often quicker and provides its own set of insights.
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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
vAve = (25m/s + -15m/s) /2
vAve = 10m/s / 2
vAve = 5m/s
With the average velocity of 5m/s and traveling at this rate for 4s the ball will travel 20m because 5m/s * 4s = 20m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
vf = 25m/s + -10m/s * 6s
vf = 25m/s - 60m/s
vf = -35m/s
vAve = (v0 + vf) / 2
vAve = (25m/s + -35m/s) / 2
vAve = -10m/s / 2
vAve = -5m/s
The ball will be -30m at the end of the sixth second because the average velocity is -5m/s and at this rate for 6 seconds, so -5m/s * 6s = -30m.
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&#This looks good. See my notes. Let me know if you have any questions. &#