Phy231
Your 'cq_1_8.1' 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:
25m/s+(-10m/s^2*1s)=15m/s
• What will be its velocity at the end of two seconds?
answer/question/discussion:
25m/s+(-10m/s^2*2s)=5m/s
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
25m/s+5m/s=30m/s/2=15m/s
• How far does it therefore rise in the first two seconds?
answer/question/discussion:
15m/s*2s=30m
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion:
25m/s+(-10m/s^2*3s)=-5m/s
25m/s+(-10m/s^2*4s)=-15m/s
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
25m/s+(-10m/s^2*Xs)=0m/s
Xs=25m/s/10m/s^2=2.5s
`ds=(25m/s+0m/s)/2=12.5m/s*2.5s=31.25m
• 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:
25m/s+(-10m/s^2*4s)=-15m/s
vAve=(25m/s+(-15m/s))/2=5m/s
`ds=5m/s*4s=20m
• How high will it be at the end of the sixth second?
answer/question/discussion:
25m/s+(-10m/s^2*6s)=-35m/s
vAve=(25m/s+(-35m/s))/2=-5m/s
`ds=-5m/s*4s=-20m (assuming the ball can fall past the point it was thrown, otherwise its height is 0 -- ground)
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15 minutes
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Excellent work.