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course phy 121
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:
v0 = 25m/s
a = 10m/s^2
`dt = 1s
Vf = 25m/s + (-10m/s^2)(1s)
Vf=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)
vf= 5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
15m/s + 5m/s / 2 = 10m/s = vAve
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Good, but the ball starts the 2-second interval at 25 m/s, not 15 m/s.
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How far does it therefore rise in the first two seconds?
answer/question/discussion:
5m/s - 0m/s = 5m/s rise
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5 m/s is a velocity, which cannot be an answer to the question 'how far'.
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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:
3seconds:
vf = 25m/s - 10m/s^2(3s)
vf = -5m/s
4seconds:
vf = 25m/s - 10m/s^2(4s)
vf= -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:
At two and a half seconds the ball has reached its maximum height, than being
2.5s(10m/s) = 25m
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2.5 seconds is correct.
However the average velocity for that 2.5 seconds isn't 10 m/s.
What is the initial velocity, what is the final, so what is the average velocity?
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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 = -15m/s +25m/s / 2
vAve = 5m/s
5m/s(4s) = 20m
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How high will it be at the end of the sixth second?
answer/question/discussion:
Vf = 25m/s + (-10m/s^2)(6s)
Vf = -35m/s
-35m/s + 25m/s / 2 = -5m/s = vAve
-5m/s(6s) = -30m
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Good answers to most questions.
A few still need correction.
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