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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.
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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: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
vf = 25m/s + (-10m/s^2)(1s)
vf = 25m/s - 10m/s
vf = 15 m/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 = 25m/s - 20m/s
vf = 5 m/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 = 30m/s / 2
vAve = 15m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = ‘ds/’dt
‘ds = vAve * ‘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: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
vf = 25m/s + (-10m/s^2)(3s)
vf = 25m/s - 30m/s
vf = -5 m/s
vf = v0 + a ‘dt
vf = 25m/s + (-10m/s^2)(4s)
vf = 25m/s - 40m/s
vf = -15 m/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
10m/s^2 * ’dt = 25m/s
‘dt = 2.5s
vAve = (vf + v0)/2
vAve = (0 + 25m/s)/2
vAve = 12.5m/s
vAve = ‘ds/’dt
‘ds = vAve* ‘dt
‘ds = 12.5m/s * 2.5s
‘ds = 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: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
vf = 25m/s + (-10m/s^2)(4s)
vf = 25m/s - 40m/s
vf = -15 m/s
vAve = (vf + v0)/2
vAve = (-15m/s + 25m/s)/2
vAve = (10m/s)/2
vAve = 5m/s
vAve = ‘ds/’dt
‘ds = vAve * ‘dt
‘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 = 25m/s - 60m/s
vf = -35 m/s
vAve = (vf + v0)/2
vAve = (-35m/s + 25m/s)/2
vAve = (-10m/s)/2
vAve = -5m/s
vAve = ‘ds/’dt
‘ds = vAve * ‘dt
‘ds = -5m/s * 4s
‘ds = -20m
The ball is 20m below where it started or back on the ground if it was thrown from the ground.
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*#&!
&#Very good responses. Let me know if you have questions. &#
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