cq_1_211

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phy121

Your 'cq_1_21.1' 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 vertically upward and caught at the position from which it was released.

• Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?

answer/question/discussion: ->->->->->->->->->->->-> :

ignoring all other factors, the ball would be travelling at the same speed. If we use the 4th equation of motion which is vf^2=v0^2+2*a*’ds. For the release vf=0 and for the ball coming back down, the v0=0. Since the displacement is the same going up as it is going down, then 2*a*’ds up= 2*a*’ds down. Since those are the same, then vf^2 up= v0^2down.

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• What, if anything, is different in your answer if air resistance is present? Give your best explanation.

answer/question/discussion: ->->->->->->->->->->->-> :

If air resistance was present, the speed of the ball would be a lower speed when released than when it is coming back down. The air resistance would decrease its height of the ball, which would make the speeds different with respect to displacement.

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Good, but it should be clear that this will reduce the speed the ball attains coming back down.

No need for revision but see also the link below.

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Then please compare your old and new solutions with the expanded discussion at the link

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