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course PHY 121
7/29 3:30
cq_1_211#$&*
PHY 121
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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cq_1_211
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PHY 121
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
Because the acceleration is 9.8 m/s^2 from gravity, and because it is released and
caught at the same position, it should be going the same speed as when it was released.
This is because gravity is pulling on it and slowing it down to the point that it stops
momentarily and returns to its original position.
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What, if anything, is different in your answer if air resistance is present? Give your
best explanation.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
Air resistance will push against the ball the same amount whether it is going up or
coming back down. It will slow it down the same in both directions, so I think that the
initial and final velocities will still remain the same.
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10 min
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If the acceleration is the same in both directions the ball will have the same speed on
its return.
Is the acceleration the same in both directions?
####Going up the acceleration is -9.8 m/s^2. Coming back down the acceleration is 9.8 m/s^2.####
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5 min more
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That would be correct if no other forces were present.
However there is air resistance.
How will air resistance affect the acceleration going up (e.g., will it increase or decrease the magnitude of the acceleration)?
How about coming down?
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