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One approach is to NEVER write the kinematics equations with 'g' in thethe
formula--even for vertical motion. The kinematics equations that contain
'a' require a constant acceleration. IF that acceleration happens to be
acceleration due to gravity, the value can be substituted at the pointwhere
numerical values are needed. The sign of 'a' is again determined by theanswers.
solver's choice up sign convention.
a) Sketch the problem. Choose a sign convention--let's choose down as
positive.
b) Determine that the problem can be done with kinematics techniques (the
second part could also be done using energy).
c) List the knowns:
v0 = 0
a = 9.8 m/s^2
s = 2 m
d) solve for time: s = v0t + .5at^2 since v0 = 0 t = (2s/a)^.5 =
(4m/9.8m/s^2)^.5 = .64 s
e) solve for vf: vf = v0 + at = 0 + 9.8m/s^2*.64s = 6.3 m/s
You can repeat the whole thing with down as negative to show that the sign
choice is arbitrary and provides the same (or at least consistent)
kinematics
Rick
**********************************************
Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, IN 46556
rtarara@saintmarys.edu
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www.saintmarys.edu/~rtarara/
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After sketching the problem and determining that this is indeed a
problem (although the second part could be solved with energy techniquesas
well), then list thewhen
----- Original Message -----
From: "Tina Fanetti" <FanettT@QUEST.WITCC.CC.IA.US>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, September 10, 2001 1:43 PM
Subject: Re: g
That's what they are doing I think.....
I will try to emphasize keeping the same coordinate system
Tina
On the other hand, if your students want to think that v = v0 - gt
thethen
object is moving downward, and v = v0 + gt when it is moving upward,
wethat
have a problem. Once the coordinate system is established, then for
problem the acceleration has to remain fixed at +g or at -g for movementin
either direction.<<<
Tina Fanetti
Physics Instructor
Western Iowa Technical Community College
4647 Stone Ave
Sioux City IA 51102
712-274-8733 ext 1429