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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.
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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: ->->->->->->->->->->->-> :
The highest point is when velocity = 0
a = -10 m/s^2
v0 = 15 m/s
vf = 0
`ds = ?
`dt = ?
`dv = vf – v0
= 0 – 15 m/s
= -15 m/s
a = `dv / dt; `dv = (a)*(`dt); `dt = `dv / a
`dt = (-15 m/s) / (-10 m/s^2)
= 1.5 s
vAve = (v0 + vf) / 2
= (15 m/s + 0 m/s) / 2
= (15 m/s) / 2
= 7.5 m/s
vAve = `ds / `dt
`ds = (vAve)*(`dt)
`ds = (7.5 m/s)*(1.5 s)
= 11.25 m
Direct reasoning: accelerating at 10 m/s^2 it takes 1.5 seconds for velocity to decrease by 15 m/s to zero. Average velocity for that interval is 7.5 m/s, so the height is 7.5 m/s * 1.5 s = 11.25 m above the starting point.
The ball was thrown upwards from 12 m. Therefore:
12 m + 11.25 m = 23.25 m
Therefore the ball reaches its highest point at 23.25 m and it took the ball 1.5 s to reach that 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: ->->->->->->->->->->->-> :
vf = ?
a = -10 m/s^2
v0 = 15 m/s
ds = 23.25 m + 23.25 m + 12 m = 58.5 m
`ds is displacement, not distance
from release to ground the displacement is -12 m.
a = (vf – v0) / `dt
vAve = (vf + v0) / 2 = `ds / `dt
`dt = (`ds)*(2 / vf + v0)
a = (vf – v0) / [(`ds)*(2 / vf + v0)]
= [(vf)^2 – (v0)^2] / (2*`ds)
vf = sqrt[[(a)*(2*`ds)] + (v0)^2]
= sqrt[[(10 m/s^2)*(2*58.5m)] + (15 m/s)^2]
= sqrt[1170 m^2/s^2 + 225 m^2/s^2]
= sqrt[1395 m^2/s^2]
= 37.3497 m/s
`dt = (`ds)*(2 / vf + v0)
= (58.5 m)*(2 / 37.3497 m/s + 15 m/s)
= 2.235 s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Speed = change in distance / change in time = vAve
vAve = 5 m/s
v0 = 15 m/s
a = -10 m/s^2
(vAve)*(`dt) = `ds
`dt = (`ds) / vAve)
vAve = (v0 + vf) / 2
(2)*vAve = v0 + vf
vf = (2*vAve) – v0
a = (vf – v0) / `dt
a = ((2*vAve) – v0 – v0) / `dt
a = ((2*vAve) – 2*v0) / `dt
(a)*(`dt) = ((2*vAve) – 2*v0)
`dt = ((2*vAve) – 2*v0) / (a)
`dt = ((2*5 m/s) – (2*15 m/s)) / -10 m/s^2
`dt = (-20 m/s) / (-10 m/s^2)
`dt = 2 s
Very good reasoning, though this can be done more simply.
The speed is 5 m/s when the velocity is 5 m/s or -5 m/s.
Starting at 15 m/s the speed will decrease to 5 m/s in 1 s, and to -5 m/s in another second.
You can use the second equation vf = v0 + a `dt, with v0 = 15 m/s and a = -10 m/s^2. `dt = (vf - v0) / a. Solve for vf = +5 m/s, then for vf = -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: ->->->->->->->->->->->-> :
vf = ?
a = 10 m/s^2
v0 = 15 m/s
ds = 20 m
a = (vf – v0) / `dt
vAve = (vf + v0) / 2 = `ds / `dt
`dt = (`ds)*(2 / vf + v0)
a = (vf – v0) / [(`ds)*(2 / vf + v0)]
= [(vf)^2 – (v0)^2] / (2*`ds)
vf = sqrt[[(a)*(2*`ds)] + (v0)^2]
= sqrt[[(10 m/s^2)*(2*20m)] + (15 m/s)^2]
= sqrt[400 m^2/s^2 + 225 m^2/s^2]
= sqrt[640 m^2/s^2]
= +-25.298 m/s
Whichever direction you choose as positive, a and `ds will have opposite signs.
20 m above the ground is 8 m above the starting point. So the displacement `ds for this interval is only 8 m.
`dt = (`ds)*(2 / vf + v0)
= (20 m)*(2 / 25.298 m/s + 15 m/s)
= 0.993 s
= (20 m)*(2 / (-25.298 m/s) + 15 m/s)
= 3.884 s
Therefore the clock times are 0.993 s and 3.884 s.
`ds = ?
v0 = 15 m/s
vf = 0 m/s
vAve = (15 m/s + 0 m/s) / 2 = 7.5 m/s
`ds = (vAve)*(`dt)
`ds = (7.5 m/s)*(6 s) = 45 m
45m – 12 = 33 m
Therefore in 6 seconds the ball will be at a height of 33m
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Around 30 - 45 min
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Excellent reasoning, but your did have some errors with your assumed quantities (e.g., you need to be very careful with signs, and you had a couple of errors related to displacement). You should submit a revision on this one.
There are also cases where simpler reasoning could have been used.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
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