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Phy 231
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: ->->->->->->->->->->->-> sion:
For the upward movement, v0=15m/s, vf=0m/s, a=-10m/s^2
a=(vf-v0)/`dt
-10m/s^2=(0m/s-15m/s)/`dt
`dt=(-15m/s)/(-10m/s^2)
`dt=1.5s
It takes 1.5 seconds to reach its peak.
During that time, it travels 1.5s * vAve, or 1.5s*(vf+v0)/2, or 1.5s*(15m/s)/2= 11.25m.
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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: ->->->->->->->->->->->-> sion:
For the downward movement, `ds=-11.25m-12m=-23.25m, v0=0m/s, a=-10m/s^2
`ds=v0`dt+.5a`dt^2
-23.25m=0+.5(-10m/s^2)(`dt^2)
-23.25m=-5m/s^2(`dt^2)
4.65m*s^2/m=`dt^2
`dt=sqrt(4.65s^2)
`dt=2.16s
It takes 2.16s to get from its peak to the ground, plus the 1.5s it took to reach its peak, so the total airtime is 3.66s.
a=`dv/`dt
-10m/s^2=`dv/2.16s
-21.6m*s/s^2=`dv
`dv=-21.6m/s
Velocity is -21.6m/s, so it is going 21.6m/s in the downward direction when it lands.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> sion:
If it starts at 15m/s and slows by 10m/s, it will be at 15m/s-10m/s =5m/s after the first full second.
It will be at 5m/s-10m/s =-5m/s after another full second.
So the speed of the ball will be at 5m/s (in two different directions) at clock time 1s and clock time 2s.
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Going from 5 m/s to -5 m/s during an interval, with velocity changing at a constant rate, the velocity will be zero at the halfway point of the interva. This occurs at clock time 1.5 seconds, halfway between 1 s and 2 s.
Note that all this depends on the linearity of the velocity function, which is equivalent to the constancy of the acceleration function.
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It would also have been possible to conclude that, changing at -10 m/s every second, the velocity would take 1.5 s to decrease from 15 m/s to 0 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: ->->->->->->->->->->->-> sion:
a=-10m/s^2
v0=15m/s
`ds=(20m-12m)=8m
vf^2=v0^2+2a`ds
vf=sqrt[(15m/s)^2+2(-10m/s^2)(8m)]
vf=sqrt(225m^2/s^2-160m^2/s^2)
vf=sqrt(65m^2/s^2)
vf=+-8.1m/s
The ball will be at 20 meters above the ground when its velocity is +8.1m/s or -8.1m/s
If vf is +8.1m/s, vAve is (15m/s+8.1m/s)2, or 11.6m/s
vAve=`ds/`dt
11.6m/s=8m/`dt
`dt=.69s
If vf is -8.1m/s, vAve -s (15m/s-8.1m/s)2, or 3.5m/s
3.5m/s=8m/`dt
`dt=2.3s
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Good.
Checking for consistency:
In .69 s the velocity will have changed by -6.9 m/s, to 8.1 m/s. Velocity will have averaged (15 m/s + 8.1 m/s) / 2 - 11.15 m/s, which multiplied by .69 s gives you about 8 meters. Added to the 12 m initial height, this gives you the 20 m.
Similar reasoning will work with the 2.3 s interval. It would be instructive to work this out.
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As another alternative, using the third equation of motion
`ds = v0 `dt + 1/2 a `dt^2
you could solve for `dt, given `dx, v0 and a.
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The ball will be at 20 meters above ground at .69s and 2.3s.
a=-10m/s^2
v0=15m/s
`dt=6s
a=(vf-v0)/`dt
-10m/s^2*6s=vf-15m/s
-60m/s=vf-15m/s
vf=-45m/s
vAve=`ds/`dt
(15m/s-45m/s)/2=`ds/6s
-15m/s*6s=`ds
-75m=`ds
If it were possible, the ball would be 75 meters below its starting point. But, the starting point was only 12m off the ground, so the ball will have already landed.
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&#Good responses. See my notes and let me know if you have questions. &#