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Phy 121
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: ->->->->->->->->->->->-> :
aAve = (vf - v0) / `dt
- 10 m/s/s = ( 0 m/s - 15 m/s ) / `dt
`dt * -10 m/s/s = -15 m/s
`dt = 1.5 s
# vAve = `ds / `dt
( 15 m/s + 0 m/s) / 2 = `ds / 1.5 s
7.5 m/s * 1.5 s = `ds
11.25 m = `ds
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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: ->->->->->->->->->->->-> :
aAve = ( vf - v0) / `dt
-10 m/s/s = (vf - 15 m/s) / `dt
`dt = (vf - 15 m/s) / -10 m/s/s
# (vf + v0) / 2 = `ds / `dt
(vf + 15 m/s) / 2 = -12 m / (vf - 15 m/s) / -10 m/s/s
(vf + 15 m/s) = -24 m * (-10 m/s/s / (vf - 15 m/s)
(vf + 15 m/s) * (vf - 15 m/s) = 240 m^2/s^2
vf^2 - 225 m^2/s^2 = 240 m^2/s^2
vf^2 = 465 m^2 / s^2
sqrt(vf^2) = sqrt(465 m^2 / s^2)
vf = 21.56 m/s This is how fast the ball is going when it hits the ground.
-10 m/s/s = (21.56 m/s - 15 m/s) / `dt Finding the time
`dt = (21.56 m/s - 15 m/s) / -10 m/s/s
`dt = .656 s
I think this is time time taken for the ball to fall from the 12 ft to the ground.
I must also consider the time taken for the ball to rise above the 12 ft and for it to fall back to the 12 ft mark.
- 10 m/s/s = ( 0 m/s - 15 m/s ) / `dt
`dt * -10 m/s/s = -15 m/s
`dt = 1.5 s
- 10 m/s/s = ( 15 m/s - 0 m/s ) / `dt
`dt * -10 m/s/s = 15 m/s
`dt = 1.5 s
# 1.5 s + 1.5 s + .656 s = 3.656 s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
-10 m/s/s = (5 m/s - 15 m/s) / `dt
`dt * -10 m/s/s = -10 m/s
`dt = 1 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: ->->->->->->->->->->->-> :
20 meters - 12 meters = 8 meters high
vf^2 = v0^2 + 2 a `ds
vf^2 = 15 m/s^2 + 2 * -10 m/s/s * 8 m
vf^2 = 225 m^2/s^2 -160 m^2/s^2
sqrt(vf^2) = sqrt(65 m^2/s^2)
vf = 8.06 m/s
vAve = `ds / `dt
(15 m/s + 8.06 m/s) / 2 = 8 m / `dt
11.53 m/s * `dt = 8m
`dt = .69 s
How high will it be at the end of the sixth second?
aAve = (vf - v0) / `dt
-10 m/s/s = ( vf - 15 m/s) / 6 s
-60 m/s = vf - 15 m/s
-45 m/s = vf
vAve = `ds / `dt
(-45 m/s + 15 m/s) / 2 = `ds / 6 s
-15 m/s * 6 s = `ds
-90 m = `ds
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Good, but you don't really need to break things down into subintervals.
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