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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_0=25 m/s
a=10m/s^2
1 second traveling 15m/s (25m/s-10m/s=15m/s)
2 seconds=5 m/s (25 m/s-2(10m/s)=5m/s)
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve=(15 m/s+25m/s)/2 =20m/s
'ds=vAve
'dt=20m/s*1 sec=20 m
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Your answers seem to be out of order with the questions, but this is the correct result for the position at the end of the first second.
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??? I am a little confused on this problem, I think I did it right????
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
(25m/s+5m/s)/2 =15 m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve=(15 m/s +25 m/s)/2
vAve=15 m/s
'dt=2 sec
'ds=vAve*'dt
'ds=15m/s* sec=30 m
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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 = -5m/s, after 4 seconds = -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: ->->->->->->->->->->->-> :
Rest=0m/s
To go from 25 m/s to 0 m/s
acceleration=-10 m/s^2
So 2.5 sec*(-10 m/s^2)= -25 m/s
It takes 2.5 seconds to get to rest.
Avg Velocity= (25 m/s+0 m/s)/2=12.5 m/s
Avg Velocity= 12.5m/s
'ds=12.5 m/s*2.5 sec
'ds=31.2 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: ->->->->->->->->->->->-> :
Velocity after 4 sec:
25 m/s-(4 *10 m/sec)= -15 m/sec
'ds=vAve*'dt
'ds=(25m/s+(-15 m/sec)/2)* 'dt
'ds=5 m/s * 4 sec
'ds= 20 m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Velocity after 6 sec:
25 m/s-(6*10m/sec)= -35 m/sec
'ds=vAve*'dt
'ds=(25m/s+(-35 m/sec)/2)* 'dt
'ds=-5 m/s * 6 sec
'ds= -30 m
??? How do I have a negative 'ds in meters???
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The object rises to its maximum height, then begins falling, falling faster and faster. If nothing (like, for example, the ground) gets in its way, at the end of 6 seconds it will be 30 m below its starting point.
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45 min
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&#Good responses. See my notes and let me know if you have questions. &#