Phy 201
Your 'cq_1_08.1' 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 upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> : I would say after one second the new velocity of the ball would be 15m/s. I figured this because if gravity is forcing the ball down 10m/s every second, the velocity of the ball is going to decrease by 10 after that first second.
• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> : Using the same logic as above, I get a new velocity of 5m/s.
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> : I determined the vAve = 15m/s. I got this by using vAve = (v0 + vf)/2, so (25m/s + 5m/s) / 2 = 15m/s.
• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> : It appears as though the ball was tossed up 30meters during those first 2 seconds. Using ‘ds = vAve * ‘dt, I get (15m/s * 2 seconds) = ‘ds = 30 meters.
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> : Using the same logic as the first two questions, the velocity of the ball after 3 seconds in the air would be –5m/s and –15m/s after 4 seconds in the air. I believe this represents the ball falling back down to earth.
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> : At the 2.5 second interval, I believe the ball has reached it’s maximum height. I also get a max height of 31.25 meters. I reasoned this because the given v0 = 25m/s. If the ball has reached it’s max height, it must be moving 0m/s, or vf = 0m/s. From that I get my vAve = 12.5m/s, and using ‘ds = vAve *’dt, I get the ‘ds = 31.25 meters.
• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> : The vAve after 4 seconds is 5m/s. Since the v0 = 25m/s and the vf = -15m/s, (v0 + vf)/2 gives me a vAve of 5m/s.
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> : Well after 5 seconds, the ball should be back to the original stop where you tossed it. After 6 seconds, I guess the ball would be at a ‘ds of – 30 meters. I used ‘ds = -5m/s * 6 seconds to get that, but I have no idea if that is right or if a distance of -30 meters even exists.
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About 18 minutes
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&#This looks very good. Let me know if you have any questions. &#