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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
&#Your work looks very good. Let me know if you have any questions. &#