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phy 231
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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: ->->->->->->->->->->->-> :
You can find the answer either by using the motion equations or reasoning. Reasoning tells you that after 1 second, with a positive velocity of 25 m/s and a negative acceleration of 10 m/s^2 that the final velocity would slow down to 15 m/s. That can be checked for vf = v0 + a*`dt = 25 m/s - 10m/s*1 = 15 m/s.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
Using reasoning again we know that after another second the velocity will slow down from 15m/s to 5 m/s because of the -10 m/s^2 acceleration. vf = 25 m/s - 10m/s/s*2s = 5 m/s.
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve for first two seconds would be = (25 m/s + 5 m/s)/2 = 15 m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
You can calculate it two ways: vAve * `dt = 15 m/s * 2 s = 30 m
Or `ds = v0*`dt + ½ a*`dt^2 = 25 m/s*2s + ½(-10m/s/s)(2s)^2 = (50 - 20) = 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: ->->->->->->->->->->->-> :
Using the same reasoning we continue to subtract 10m/s^2 from the velocity for each additional second. So 3 seconds would yield a velocity 0f -5 m/s and 4 seconds would yield -15 m/s. The equations show the same answer.
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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: ->->->->->->->->->->->-> :
When velocity is zero we know it has reached its maximum height. Using reasoning or the equations we can deduce that it will take 2.5 sec for the ball to reach its maximum height.
`a * `dt = `dv -10m/s^2 * 2.5s = -25cm/s
`dv + v0 = vf -25 + 25 = 0 (max height)
(vf + v0)/2 = vAve (0 + 25)/2 = 12.5 m/s
vAve * `dt = `ds 12.5 m/s * 2.5s = 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: ->->->->->->->->->->->-> :
At 4 sec, the velocity is at -15 m/s as we calculated in previous problems and the v0 is still 25 m/s.
(-15 m/s + 25 m/s) / 2 = 5 m/s
vAve * `dt = `ds 5 m/s * 4s = 20 m
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
By the 5th second the ball will be back on the ground, so at the end of the sixth second the values no longer are relevant.
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45 min
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&#Very good responses. Let me know if you have questions. &#