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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.
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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: ->->->->->->->->->->->-> :
v=v0+a*t
v=25m/s - 10m/s^2 * 1= 15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
v=v0+a*t
v=25m/s-10m/s^2 * 2 = 5m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= v0+v2/ 2
vAve= 25m/s + 5m/s / 2 = 15m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
15m/s * 2s = 30m
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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: ->->->->->->->->->->->-> :
v=v0+a*t
v= 25m/s -10m/s^2 * 3s= -5m/s
v=25m/s-10m/s^2* 4s = -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: ->->->->->->->->->->->-> :
v=v0+a*t
0= 25m/s-10m/s^2 * t
-25m/s=-10m/s^2*t
2.5s = t
t=2.5 sec.
vAve= v0+v2.5/2 = 25m/s +0m/s/ 2 = 12.5m/s
12.5m/s * 2.5s = 31.25m = maximum ht
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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: ->->->->->->->->->->->-> :
vAve= v0+v4/ 2 = 25m/s + -15m/s / 2 = 5m/s
5m/s *4s= 20m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
v=v0+a*t
v=25m/s + -10m/s^2 * 6s
v= -35m/s
vAve=v0+v6 / 2 = 25m/s + -35m/s / 2 = -5m/s
-5m/s * 6s = -30m
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20min
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&#Very good responses. Let me know if you have questions. &#