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phy121
Your 'cq_1_08.2' 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_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
we want to find `ds and `dt. We are given v0 = 15m/sec a = -10 m/sec^2
and vf =0 since the moment the object reaches its highest point, it stops for an instant before falling back down to earth. So
vAve = 15+0/2 = 7.5 m/sec
a = `dv/`dt
-10m/sec^2 = -15m/sec/`dt
-10m/sec^2`dt = -15m/sec
`dt = 1.5 sec
`dv = 0-15 = -15m/sec
v= `ds/`dt
7.5m/sec = `ds/1.5sec
`ds= 11.25 meters.
The object rises 11.25 meters and it takes 1.5 seconds to reach this high point.
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
we are told that v0 = 15m/sec, a = -10m/sec^2 and that `ds is -12 m since the object ended up 12 m lower than it started.
So vf^2 = v0^2 +2*a*`ds
vf^2 = 15m/sec^2 + 2 * -10m/sec^2 * -12m
vf^2 = 465
vf = 21.6 m/sec
`dv = 21.6 - (-15) = 36.6 m/sec
a = `dv/`dt
-10m/sec^2 = 36.6 m/sec / `dt
`dt = -3.66 sec or 3.66 seconds.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
we want to know when
vf = 5m/sec
v0 = 15m/sec
a = -10 m/sec
a = `dv / `dt
`dv = 5-15 = -10m/sec
-10m/sec = -10m/sec/`dt
`dt = 1sec
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Speed is the magnitude of the velocity, so the speed will be 5 m/s when the velocity is +5 m/s, and when the velocity is -5 m/s.
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
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We want to find `dt when the ball is 20 meters above the ground
The original position of the ball is 12 m above the ground so this would be a `ds = 20/12 = 8m
Given `ds = 8m, v0 = 15m/sec and a = -10 m/sec^2
Vf^2 = v0^2 + 2*a*’ds
Vf^2 = (15m/sec)^2 + 2*-10m/sec^2 * 8m
vf= 8.06 m/sec
so ‘dv = 8.06 m/sec - 15 m/sec = -6.94 m/sec
so a = `dv /`dt gives -10m/sec^2 = -6.94m/sec /`dt so `dt = .694 seconds
at the end of 6 seconds, `dt = 6seconds, v0 = 15m/sec a = -10m/sec and we want to find `ds
vf = 15m/sec + -10m/sec^2 * 6
vf = -45 m/sec
`dv = -45m/sec - 15m/sec = -60 m/sec
vAVe = - 45+ 15 = -30 / 2 = -15 m/sec
`ds = vAve * `dt
= -15m/sec * 6sec =-90 meters + 12m above ground intitially = -78 meters
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30 minutes
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2:15 pm 12/13/13
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