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PHY 201
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.
** CQ_1_08.1_labelMessages **
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, a = -10 m/s^2, 'dt = 1, find 'ds after 1 second.
`ds = v0 `dt + .5 a `dt^2
'ds = (25 m/s)(1 s) + 0.5(-10 m/s^2)(1 s)^2 = 20 m
vf = 25 m/s + (-10 m/s^2)(1 s) = 15 m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a * `dt
vf= 25 m/s + (-10 m/s^2)(2 s) = 5 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (vf + v0) / 2 = (5 + 25) / 2 = 15 m/s.
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = v0 `dt + .5 a `dt^2
'ds = (25 m/s)(2 s) + 0.5(-10 m/s^2)(2 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: ->->->->->->->->->->->-> :
vf = v0 + a * `dt = 25 m/s + (-10 m/s^2)*3 s = -5 m/s (changing direction downward)
vf = v0 + a * `dt = 25 m/s + (-10 m/s^2) * 4 s = -15 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: ->->->->->->->->->->->-> :
`ds = (vf + v0) / 2 * `dt
at 2 seconds it reached 30 m
@3 seconds 'ds = (-5 m/s + 25 m/s) / 2 * 3 s = 30 m
@4 seconds 'ds = (-15 m/s + 25 m/s) / 2 * 4 s = 20 m
I would surmise from the data that at 2 seconds, the ball reaches its maximum height at 30 m and then begins to fall.
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At the 2-second mark it is still rising at 5 m/s, so it hasn't yet reached its maximum height.
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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 of the first four seconds would be 5 m/s
(25 m/s + 15 m/s + 5 m/s + -5 m/s + -15 m/s) / 5 = 5 m/s
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = v0 `dt + .5 a `dt^2
'ds = (25 m/s)(6 s) + 0.5(-10 m/s^2)(6 s)^2 = 120 m, I don't believe this, it would appear the ball is coming back down at 4 seconds, levels out at 3 seconds at 30 m.
vf = v0 + a * `dt = 25 m/s + (-10 m/s^2)(6 s) = - 35 m/s
`ds = (vf + v0) / 2 * `dt = (-35 m/s + 25 m/s)/2 * 6 s = -30 m, or in this case 0 m. This makes the most sense to me.
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Good, but see my note about the maximum height. It can't be 30 meters.
Reason it out, and if you can't get the correct answer, which is 32.5 meters, let me know.
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