Phy 201
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: aAve = ‘dv/’dt so -10m/s^2 = vf-25m/s/1s. -10m/s = vf – 25m/s. -10m/s + 25m/s = 15m/s
• What will be its velocity at the end of two seconds?
answer/question/discussion: -10m/s^2 = vf-25m/s/2s. -20 m/s = vf-25m/s. 5m/s = vf
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: vAve = (vf + vo)/2 (5m/s + 25m/s)/2 is 30/2 = 15m/s
• How far does it therefore rise in the first two seconds?
answer/question/discussion: ‘ds = vAve * ‘dt. ‘ds = 15m/s * 2s ‘ds = 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: -10m/s^2 = vf-25m/s/3s. -10*3 = v-25m/s -30m/s +25m/s = -5m/s
-10m/s^2 = v-25m/s/4s. -10m/s *4s=vf-25m/s -40m/s +25m/s =vf -15m/s = vf
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: hmax = v0^2/2g so 25m/s^2/2(10) = 625/20=31.25 m
• 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: vAve = (-15m/s +25m/s)/2 vAve = 5m/s
‘ds = vAve(‘dt) so 5m/s(4s)is’ds = 20m
• How high will it be at the end of the sixth second?
answer/question/discussion: -10m/s^2 = vf-25m/s/6s. -10*6s = vf-25m/s -60+25=vf -35m/s =vf so ‘ds = vAve(‘dt) ‘ds = -35m/s +25m/s /2. -5*6 so ‘ds = -30m
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50min
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Excellent work on all questions.