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phy121
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: ->->->->->->->->->->->-> :
we are told v0 = 25m/sec a = -10 m/sec^2 (because it is downward) and we want vf after `dt = 1sec
vf = v0 + a * `dt
= 25m/sec + (-10m/sec^2)*1
= 15 m/sec
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
so now we have v0 = 25m/sec a = -10 m/sec^2 and `dt = sec
so vf =25m/sec + -10m/sec^2 * 2sec
= 5m/sec
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAVe = (vf+v0)/2 for the first 2 seconds, v0 = 25m/sec and vf = 5 m/sec
vAve = (25+5)/2 = 15 m/sec
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = vAVe * `dt
= 15m/sec * 2sec
= 30 meters
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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: ->->->->->->->->->->->-> :
at `dt = 3 vf = 25m/sec + -10m/sec^2 * 3sec = -5m/sec
at `dt = 4 25m/sec + -10m/sec^2 * 4sec = -15 m/sec
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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: ->->->->->->->->->->->-> :
the max height would be when vf = 0 because that is the instant when the ball begins its fall back to earth so
if vf = 0
a = -10m/sec^2
v0 = 25m/sec
we want to find `ds and `dt
vf = v0 + a*`dt
0 = 25 + -10*`dt
-10`dt= -25
`dt = 2.5 sec
Vave = (25+0)/2 = 12.5 m/sev
`ds= vave*’dt
= 12.5m/sec * 2.5 sec=
’ds =31.25 meters
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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: ->->->->->->->->->->->-> :
avgVel = (-15+25)/2 = 5 m/sec
`ds = 5m/sec * 4 sec = 20 m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
so we have `dt = 6sec
vf = v0 + a*`dt
= 25 + (-10 * 6) =-35 m/sec
Vave then is -35 +25 = -10/2 = -5 m/sec
So `ds = vAve *`dt = -5m/sec * 6 sec = -30 meters
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*#&!
&#Very good responses. Let me know if you have questions. &#