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Phy 121
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: ->->->->->->->->->->->-> :
v=25m/s-10m/s(1s)=15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
v=25m/s-10m/s(2s)=5m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
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: ->->->->->->->->->->->-> :
‘ds=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: ->->->->->->->->->->->-> :
v0=25m/s
a=10m/s^2
‘dt=3s
vf=25m/s-10m/s^2(3s)=-5m/s
v0=25m/s
a=10m/s^2
‘dt=4s
vf=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: ->->->->->->->->->->->-> :
Ball reaches maximum height at vf = 0 and it has risen 31m.
0m/s = 25m/s + (-10m/s^2)’dt
-25m/s = (-10m/s^2)’dt
‘dt=-25m/s/(-10m/s/s) =2.5s
‘ds = 25m/s / 2 * 2.5s
‘ds = 12.5m/s * 2.5s
‘ds = 31m
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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: ->->->->->->->->->->->-> :
vf = -15m/s
v0 = 25m/s
vAve = (25m/s - 15m/s) /2 = 5m/s
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = 25m/s
a = -10m/s/s
‘dt = 6s
vf = 25m/s -10m/s/s *6s
vf = 25m/s - 60m/s = -35m/s
‘ds = (25m/s - 35m/s) / 2 * 6s
‘ds = -5m/s * 6s
‘ds = 30m
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*#&!
&#Very good responses. Let me know if you have questions. &#