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PHY 121
Your 'cq_1_08.2' 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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cq_1_082
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PHY 121
Your 'cq_1_08.2' 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 at 15 meters / second from a height of 12 meters above the
ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2%
of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
I believe that it will go 24.5 m. This is because during the first second, the initial
velocity is 15 m/s and it loses 10 m/s due to gravity. This makes the final velocity of
this segment 5 m/s. This averages to 10 m/s, so it is up 12 + 10 = 22. Then it starts
this part at 5 m/s, but the gravity pulls at 10 m/s, so it will only go up half-way.
But the average velocity for this segment is 5/2, or 2.5. That makes the total 24.5 m.
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Very good thinking, but there is an error. From +5 m/s to -5 m/s the average velocity is
0 (as it must be since the ball is at the same height at both times).
The ball reaches its max height at the instant it comes to rest. Thus:
It takes 1.5 sec for the velocity to reach zero, during which ave velocity is (15 m/s +
0 m/s) / 2 = 7.5 m/s.
7.5 m/s * 1.5 s = 12.25 meters.
*@
How fast is it then going when it hits the ground, and how long after the initial toss
does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
It's going 25 m/s and it is 4 seconds after the original toss. All this is from my
picture, not math.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
It will be at 5 m/s at 1 second in and at 2 seconds in.
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At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
I guess I am hopelessly lost at this point. Now I'm trying math, but it's letting me
down. I know that I'm missing something that probably simple. But I can't seem to find
it. I don't usually give up, but I'm flummoxed!
&&&&I tried to solve the problem before I looked at the answer. My primary confusion
was that I was thinking of the process in two steps: from 15 m/s to 0 m/s and then from
0 m/s 23.25 meters below. I understand the answer and did work the math on my own as I
looked at the explanation, but it will take a while for me to process that -10 m/s^2
will stay -10 m/s^2 through the whole up and down process. And that the 13.25 m up and
13.25 m down cancel each other other so that you end up with a displacement of -12 m,
and finally that it it considered one trip, not two. &&&&
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You've hit the nail on the head. Those are the things that are difficult to process.
Just stating it that clearly is half the battle.
*@
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