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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: ->->->->->->->->->->->-> :
sf = 1/2a`dt^2 + v0t + s0
0m/s = 15m/s - 10m/s^2*`dt
10m/s^2*`dt = 15m/s
`dt = 15m/s / 10m/s^2
`dt = 1.5sec
Sf = 1/2(-10m/s^2)(1.5sec)^2 + 15m/s(1.5sec) + 12m
Sf = 23.25m
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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: ->->->->->->->->->->->-> :
0 = 1/2 (-10m/s^2)(t)^2 + 15m/s(t) + 12m
-12m =
@& You lost the square on one of your t's. Note that the units of -5m/s^2(t) + 15m/s(t) are not compatible; the two terms would have different units.*@
-12m = t(-5m/s + 15m)
-12m / 10m/s = 1.2s
Vf = 15m/s + -10m/s^2 * 1.2
Vf = 3m/s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
5m/s = 15m/s + -10m/s^2 * t
-10m/s = -10m/s^2 * t
T = 1sec
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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: ->->->->->->->->->->->-> :
20m = 1/2(-10m/s^2)(t)^2 + 15m/s(t) + 12m
8m = (-10m/s^2)(t)^2 + 30m/s(t)
I’m having trouble figuring out what to do from here. When I factor the equation, a T is still in the parentheses and I’m not exactly sure what to do from there. I might be approaching the problem incorrectly.
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25 minutes
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@& You had an error. It will be easy to correct, but there's a lesson in that mistake so it's worth your time to correct it.
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