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?
At the same speed.
• What, if anything, is different in your answer if air resistance is present? Give your best explanation.
I think that it will still be the same if air resistance is present. Air resistance is similar to gravity, in that it will affect it while it travels in either direction.
Air resistance will enhance the slowing effect of gravity on the rising ball, which will as a result not rise as far. As a result of the decreased maximum height, the falling ball won't drop as far, resulting in a lesser final speed.
Air resistance will then oppose the speeding up of the falling ball, so that the final velocity will be even less.
For a dense ball tossed gently upward, the effect of air resistance will be small, perhaps negligible with respect to the accuracy of our instruments. If the ball is less dense, and/or speeds are greater, air resistance will have a greater effect.
In terms of energy conservation, assuming 'still' air (i.e., no wind, no rising and falling of air), air resistance acts always in the direction opposite motion and therefore does negative work on the object. PE doesn't change between the beginning and the end of the interval, so the result is a lesser KE at the end than at the beginning.
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5 mins.
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&#Good responses. See my notes and let me know if you have questions. &#