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Phy 121
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.
** CQ_1_08.1_labelMessages **
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: ->->->->->->->->->->->-> scussion:
after one second it will have lost 10m/s from its intitial velocity so it will be 15m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
It will be 5m/s 25m/s - 10m/s(2)
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
(v0 +vf)/2 = (25m/s + 5m/s)/2 = 30m/s/2 = 15m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
traveling 15m/s for 2 seconds it would rise 30m.
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What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
The velocity will be -5m/s and -15m/s respectively assuming the direction of motion is measured away from the ground.
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At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
It reaches maximum height when its velocity is 0m/s which would occur at 2.5 seconds. It will have risen vAVe *2s=(v0+vf)/2 *2s = 25ms/2 *2 s = 25m
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Your reasoning is correct throughout; but as you say the interval is 2.5 seconds. You used 2 seconds when calculating your displacement.
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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: ->->->->->->->->->->->-> scussion:
vAve = (25m/s + -15m/s)2 = (10m/s)/2 = 5m/s
5m/s * 4 m = 20 m
Seems like it would be have fallen more than 5 meters in 1.5 seconds. Hmmm
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Good. See my previous note. The ball rose higher than 25 m.
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
After six seconds the velocity will be -35m/s
vAve will be( 25m/s + -35m/s)/2 = (-10m/s)/2 = -5m/s
-5m/s * 6 s = -30m. The ball we be 30 meters lower than it was when it was released. If it was released from a height of 40m, it will now be 10m above the ground. If it was released from a height of 2 meters, it will already have landed.
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&#Good responses. See my notes and let me know if you have questions. &#