PHY 121
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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?
25m/s / 10m/s^2 = 2.5m/s
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• What will be its velocity at the end of two seconds?
2.5m/s * 2 = 5.0m/s
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• During the first two seconds, what therefore is its average velocity?
5.0m/s * 2s = 10m/s
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• How far does it therefore rise in the first two seconds?
25m/s – 10m/s = 15m/s
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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?
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
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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?
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• How high will it be at the end of the sixth second?
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30 min
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&#Please compare your solutions with the expanded discussion at the link
Solution
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