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phy121
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: ->->->->->->->->->->->-> :
vf= v0+a*'dt
vf= 25m/s+(-10m/s^2)*1s
vf= 15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf= v0+a*'dt
vf= 25m/s+(-10m/s^2)*2s
vf= 5m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= (vf+v0)/2
vAve= (5m/s+25m/s)/2
vAve= 15m/s during the first 2 seconds
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = (vf + v0) / 2 * `dt
‘ds= 15m/s*2s
‘ds= 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: ->->->->->->->->->->->-> :
3 seconds:
vf= v0+a*'dt
vf= 25m/s+(-10m/s^2)*3
vf= -5m/s
4 seconds:
Vf= v0+a*’dr
Vf= 25m/s+(-10m/s^2)*4
Vf= -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: ->->->->->->->->->->->-> :
vf= v0+a*'dt
(0m/s-25m/s)/-10m/s^2=’dt
‘dt= -25m/s / -10m/s^2
‘dt= 2.5s
`ds = (vf + v0) / 2 * `dt
‘ds= (0m/s+25m/s)/2*2.5s
‘ds= 31.25m
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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= (vf+v0)/2
vAve= (-15m/s+25m/s)/2
vAve= 5m/s
`ds = (vf + v0) / 2 * `dt
‘ds= (-15m/s+25m/s) /2*4s
‘ds= 5m/s*4s
‘ds= 20m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf= v0+a*'dt
vf= 25m/s+(-10m/s^2)*6s
vf= -35m/s
`ds = (vf + v0) / 2 * `dt
‘ds= (-35m/s+25m/s) /2*6s
‘ds= -5m/s*6s
‘ds= -30m
Which means, the ball that was tossed, was tossed off a edge, because it ended up 30m BELOW the initial height it was thrown at.
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** **
15 minutes
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sept 26, 3:40pm
&#Very good responses. Let me know if you have questions. &#