Phy 201
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: ->->->->->->->->->->->-> :
vf = v0 + a * `dt
0 m/s = 15 m/s + (-10 m/s/s) * `dt
-15 m/s = (-10 m/s/s) * `dt
1.5 s = `dt
`ds = [(vf + v0) / 2 ] * `dt
`ds = [(15 m/s + 0 m/s) / 2] * 1.5 s
`ds = 11.25 m
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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: ->->->->->->->->->->->-> :
`ds = (11.25 m * 2) + 12 m
`ds = 34.5 m
vf^2 = v0^2 + 2 * a * `ds
vf^2 = (15 m/s)^2 + 2 (-10 m/s/s)(34.5 m)
vf^2 = 225 m^2/s^2 – 690 m^2/s^2
vf^2 = -465 m^2/s^2
vf = -21.564 m/s
vf = v0 + a * `dt
-21.564 m/s = 10 m/s + (-10 m/s/s) * `dt
-31.564 m/s = -10 m/s/s * `dt
3.156 s = `dt
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a * `dt
5 m/s = 15 m/s + (-10 m/s/s) * `dt
-10 m/s = -10 m/s/s * `dt
1 s = `dt
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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: ->->->->->->->->->->->-> :
There are two possible clock times when the ball could be 20 m above the ground:
Situation A:
vf^2 = v0^2 + 2 * a * 8m
vf^2 = (15 m/s)^2 + 2 (-10 m/s/s)(8m)
vf^2 = 225 m^2/s^2 -160 m^2/s^2
vf^2 = 65 m^2/s^2
vf = 8.06 m/s
vf = v0 + a * `dt
8.06 m/s = 15 m/s + (-10 m/s/s) * `dt
-6.94 m/s = -10 m/s/s * `dt
.694 s = `dt
Situation B:
vf^2 = v0^2 + 2 * a * 26.5 m
vf^2 = (15 m/s)^2 + 2 (-10 m/s/s)(26.5 m)
vf^2 = 225 m^2/s^2 – 530 m^2/s^2
vf^2 = -305 m^2/s^2
vf = -17.46 m/s
vf = v0 + a * `dt
-17.46 m/s = 15 m/s + (-10 m/s/s) * `dt
-32.46 m/s = -10 m/s/s * `dt
3.246s = `dt
At the end of the sixth second the ball will be on the ground, having arrived there at 3.156 seconds.
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20 minutes
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&#Very good work. Let me know if you have questions. &#