Your 'cq_1_8.1' 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?
v0=25m/s; a= -10m/s^2; ‘dt = 1s
vf=v0+a*’dt = 25m/s -10m/s^2*1s=25m/s-10m/s=15m/s
answer/question/discussion:
• What will be its velocity at the end of two seconds?
vf=v0+a*’dt = 25m/s -10m/s^2*2s=25m/s-20m/s=5m/s
answer/question/discussion:
• During the first two seconds, what therefore is its average velocity?
vAve=(vf+v0)/2=(5m/s+25m/s)/2=30m/s/2=15m/s
answer/question/discussion:
• How far does it therefore rise in the first two seconds?
The average velocity rises from 0m/s to 20m/s in one second to 15m/s in two seconds.
Correct. How far does the object therefore rise during this interval?
answer/question/discussion:
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
Plus one additional second the velocity would be vf=v0+a*’dt = 25m/s =-10m/s^2*3s=25m/s-30m/s=-5m/s and -15m/s after an additional second
answer/question/discussion:
25m/s =-10m/s^2*3s=25m/s-30m/s=-5m/s implies that 25 m/s = -5 m/s.
The = sign means exactly what it says. Equality is a transitive property. The = sign does not indicate 'train of thought'.
In this case your intent is clear and correct. However in the future be sure you confine your use of the = sign to its correct definition. You're not likely to confuse me, though I do want to see things in readable format, but this sort of usage would be very confusing to other readers and, if you are reviewing for a test, to yourself.
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
Max height occurs when velocity = 0m/s
`ds= (v^2-v0^2)/2A = (0-(25m/s )^2)/2(-10m/s^2)=31m
‘dT=-v0/a=-25m/s/-10m/s^2=2.5s
answer/question/discussion:
• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
Vf of 4s =-15m/s vAve = (25m/s+-15m/s)/2 =5m/s
‘ds =vAve *’dt=5m/s*4s =20m
answer/question/discussion:
• How high will it be at the end of the sixth second?
The ball should have reached the end of its course before six seconds
You can answer this question by assuming that acceleration remains constant, then pointing out that if the ground it solid this assumption would not be valid for the entire time interval.
If there's a deep enough hole in the ground at the right place the acceleration might remain constant for the entire interval.
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45mins
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Would like to have feedback. Thanks, Joshua
&#Good responses. See my notes and let me know if you have questions. &#