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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 **
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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: ->->->->->->->->->->->-> scussion:
v0= 25m/s
a=-10m/s^2
dt=1s
vf=unknown
Using vf=v0 + a*dt:
vf= 25m/s+ -10m/s^2*1s
vf= 25m/s + -10m/s
vf=15m/s
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Good, though this can be reasoned out directly with no need to refer to equations:
Starting at 25 m/s, with velocity changing by 10 m/s per second in the downward direction, in one second the velocity will change by 10 m/s downward. Thus the upward 25 m/s velocity will decrease to 15 m/s.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
v0= 25m/s
a=-10m/s^2
dt=2s
vf=unknown
Using vf=v0 + a*dt:
vf= 25m/s+ -10m/s^2*2s
vf= 25m/s + -20 m/s
vf= 5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
(5m/s + 25m/s)/2 = 30m/s/2 = 15m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
`ds= 25m/s*2s + .5(-10m/s^2)*2s^2
`ds= 50m + -5m/s^2*2s^2
`ds= 50m + -10m
`ds = 40m
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Direct reasoning: In 2 s the velocity decreases to 5 m/s so for the first 2 s the average velocity is 15 m/s. In two seconds, then, the ball rises 30 meters.
Your equation was not worked out correctly. See if you can spot your error.
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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: ->->->->->->->->->->->-> scussion:
v0= 25m/s
a=-10m/s^2
dt=3s
vf=unknown
Using vf=v0 + a*dt:
vf_3= 25m/s+ -10m/s^2*3s
vf_3= 25m/s + - 30m/s
vf_3=-5m/s
vf_4= 25m/s+ -10m/s^2*4s
vf_4= 25m/s + - 40m/s
vf_4=-15m/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: ->->->->->->->->->->->-> scussion:
When vf=0m/s, the ball has reached its max height which is found by setting vf =0 and solving for dt:
vf=v0 + a *dt
0m/s = 25m/s + -10m/s^2 * dt
-25m/s = -10m/s^2*dt
-25m/s / -10m/s^2 =dt
1.5 s =dt
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The ball must slow by 25 m/s, slowing as it rises at 10 m/s every second.
So it takes 2.5 seconds to slow to rest.
You have an arithmetic error in the last of your calculations.
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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: ->->->->->->->->->->->-> scussion:
vAve = (25m/s + -15m/s)/2= 10m/s/2 = 5m/s
`ds= 25m/s(4s) + .5(-10m/s^2)*4s^2
`ds= 100m + -5m/s^2*16s^2
`ds=100m - 80m
`ds=20m
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
`ds=25m/s*6s + -5m/s^2*6s^2
`ds=150m - 180m
`ds=-30m
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You have a couple of errors, and could have reasoned out all these results then compared the reasoned-out results with the results of the equations.
See my notes and let me know if you have questions.
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