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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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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