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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: ->->->->->->->->->->->-> :
Given: velocity up=25m/s, accel gravity=10m/s^2 down
Find ball velocity after 1s:
vf=v0 + a ‘dt
vf=25m/s + -10m/s^2 * 1s = 15m/s
after 1 second the ball’s upward velocity will be 15m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf=25m/s + -10m/s^2 * 2s = 5m/s
After 2 seconds the ball’s upward velocity is 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=15m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve*’dt =’ds
15m/s * 2s = 30m
It rises 30m in the first 2 seconds.
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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: ->->->->->->->->->->->-> :
velocity at +1s and +2s(3s and 4s cumulative):
vf=v0 + a ‘dt
vf=25m/s + -10m/s^2 * 3s = -5m/s
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: ->->->->->->->->->->->-> :
Max height is when vf=0
vf^2=v0^2 + 2’ds
‘ds=(vf^2-v0^2)/(2*a)
‘ds=(0^2-25m/s)/(2*-10m/s^2)=31.25m
Max height is 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: ->->->->->->->->->->->-> :
Ave velocity after 4s is:
(vf+v0)/2=(-15m/s + 25m/s)/2 = 5m/s
Height of ball after 4s is:
‘ds=v0 ‘dt + .5 a ‘dt^2
‘ds=(25m/s*4s) + (.5*-10m/s^2*(4s)^2)=20m
After 4s ball is 20m high.
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
‘ds=v0 ‘dt + .5 a ‘dt^2
‘ds=(25m/s*6s) + (.5*-10m/s^2*(6s)^2)=-30m
Looks like we went under the ground!
I’d guess all the motion stopped at ‘ds =0
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... in the event of a deep well, your solution could be useful.
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*#&!*#&!
&#Good responses. See my notes and let me know if you have questions. &#