cq_1_081

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Mth 272

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

v0 = 25 m/s

g = -10m/s^2

vf = ?

vf = v0 + (a)(t)

vf = 25 m/s + (-10 m/s^2)(1 s)

vf = 15 m/s

The velocity of the ball after one second equals: 15 m/s

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

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

vf = v0 + at

vf = 25 m/s + (-10m/s^2)(2s)

vf = 5 m/s

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

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

vf + v0 / 2 = 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: ->->->->->->->->->->->-> :

15 m/s * 2 s = `ds

`ds = 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: ->->->->->->->->->->->-> :

Velocity at 3 seconds:

Vf = 25 m/s + (-10m/s^2)(3s)

Vf = -5 m/s (three seconds)

Velocity at 4 seconds:

Vf = 25 m/s + (-10m/s^2)(4s)

Vf = -15 m/s (four seconds)

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

The Ball reaches its maximum height at final velocity of 0 m/s.

V = vo + at

0 = 25 m/s + (-10m/s^2)(t)

-25 m/s / -10 m/s^2 = t

t = 2.5 s (ball reaches its maximum position)

v^2 = vo^2 + 2a(y – y0)

y – y0 = -625 m/s / 2(-10 m/s^2)

y – y0 = 31. 25 m (ball reaches its maximum distance)

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

The average velocity for the first four seconds = 0 m/s

15 m/s + 5 m/s +(-5 m/s) + (-15 m/s) / 4 = 0 m/s

The height at the end of the fourth second:

I want to say 0 m for the fourth seconds; however, here are my results:

V^2 = v0^2 + 2a(y – y0)

225 m/s = 625 m/s + 2(-10 m/s^2)(y – y0)

y – y0 = 20 m

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

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

vf = 25 m/s + (-10 m/s^2)(6s)

vf = -35 m/s

v^2 = vo^2 + 2ª(y – y0)

1225 m/s = 625 m/s + 2(-10 m/s^2)(y-y0)

600 m/s / -20 m/s^2 = y – y0

y – y0 = - 30 meters high at six seconds

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1 hour

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9/28 3pm

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