cq_1_81

Phy 201

Your 'cq_1_8.1' 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.

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What will be the velocity of the ball after one second?

answer/question/discussion: after 1 s = 25 m/s + (-10 m/s^2)(1 s) = 15 m/s = v

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What will be its velocity at the end of two seconds?

answer/question/discussion: after 2 s = 25 m/s + (-10 m/s^2)(2 s) = 5 m/s = v

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During the first two seconds, what therefore is its average velocity?

answer/question/discussion: (5 m/s + 25 m/s) / 2 = 15 m/s

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How far does it therefore rise in the first two seconds?

answer/question/discussion: [(5 m/s)^2 - (25 m/s)^2] / 2 * 10 m/s^2 = 30 m

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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: after 3 s = 25 m/s + (-10 m/s^2)(3 s) = -5 m/s = v; after 4 s = 25 m/s + (-10 m/s^2)(4 s) = -15 m/s = v

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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: v= vo + at; t = - vo / a = 25 m/s / -10 m/s^2 = 2.5 s; y = vot + 1/2at^2 = 25 m/s* 2.5 s + 1/2(-10 m/s^2)(2.5)^2 = 31.25 m

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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: (-15 m/s + 25 m/s) / 2 = 5 m/s; y = vot + 1/2at^2 = 25 m/s* 4 s + 1/2(-10 m/s^2)(4)^2 = 20 m

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How high will it be at the end of the sixth second?

answer/question/discussion: y = vot + 1/2at^2 = 25 m/s* 6 s + 1/2(-10 m/s^2)(6)^2 = -30 m

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25 minutes

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Excellent work.