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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: ->->->->->->->->->->->-> :
vf = v0 + a * t
0 = 15 -10 * t
t = 1.5 s
y-y0 = v0 * t + 0.5 * a * t^2
y-12 = 15 * 1.5 + 0.5 * -10 * 1.5^2
y = 15 * 1.5 + (0.5 * -10 * 1.5^2) + 12
y = 23.25 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: ->->->->->->->->->->->-> :
x-x0 = v0 * t + 0.5 * a * t^2
-23.25 = 15 * t + 0.5 * -10 * t^2
t = 4.13 s
vf = v0 + a * t
vf = 15 - 10 * 4.13
vf = 26.3 m/s downward
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At what clock time(s) will the speed of the ball be 5 meters /
second?
answer/question/discussion: ->->->->->->->->->->->-> :
v = v0 + a * t
5 = 15 + -10 * t
t = 1.0 sec
@& Speed is 5 m/s if the velocity is 5 m/s or -5 m/s. You didn't consider the latter. Easily done, of course, but not as easy to remember to consider it.*@
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Good responses. See my notes and let me know if you have questions.