cq_1_081

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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: ->->->->->->->->->->->-> :

25 m/s

@& 25 m/s is the initial velocity, not the velocity 1 second later.*@

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

answer/question/discussion: ->->->->->->->->->->->-> :

15 m/s

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

answer/question/discussion: ->->->->->->->->->->->-> :

(25 m/s + 15 m/s) / 2 = 20 m/s

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

answer/question/discussion: ->->->->->->->->->->->-> :

20 m/s * 2 = 40 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: ->->->->->->->->->->->-> :

5 m/s

-5 m/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 3.5 s when the ball evens out at 0 m/s

20 m/s * 3.5 = 70 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: ->->->->->->->->->->->-> :

(25 m + 15 m + 5 m + - 5m) / 4 = 11.25 m/s

11.25 m/s * 4 s = 45 m

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

answer/question/discussion: ->->->->->->->->->->->-> :

(25m + 15 m + 5 m + -5m + -15 m + -25m) / 6 = 0 m/s

0 m/s * 6 s = 0 m

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*#&!*#&!

@& Your reasoning is good throughout; but you had the initial velocity occuring 1 second after the start, which influenced your results.*@

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No need for a revision.*@