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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Eric & Steven             \+
   Feb 24, 2000" }}{PARA 0 "" 0 "" {TEXT -1 11 "PP. 97, #12" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "re
start:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 435 "A shell of mass m = 2 \+
kg is shot straight up at  initial velocity v = 200 m/s from the groun
d.  Our task is to determine the time this shell reaches it's maximum \+
height, and what that height may be.  We assigned ground level as x = \+
0, and upwards as the postive x direction.  Two forces act in the oppo
site direction, the force due to air resistance and gravity.  Since we
 are only concerned with the trip up, we my consider the force " }}
{PARA 0 "" 0 "" {TEXT -1 199 "m * dv/dt (or dp/dt) to be positive, and
 air resistance k * v and gravity m * g to be negative.   As we know t
he shell will lose velocity, that dp/dt will be negative, this intuiti
vely makes sense.  " }}{PARA 0 "" 0 "" {TEXT -1 82 "                  \+
                                         m * dv/dt = -k * v - mg" }}
{PARA 0 "" 0 "" {TEXT -1 43 "Plugging the known constants, this become
s:" }}{PARA 0 "" 0 "" {TEXT -1 90 "                                   \+
                        2 * dv/dt = -.05 * v - 2 * 9.81" }}{PARA 0 "" 
0 "" {TEXT -1 124 "Recognizing this as a seperable equation, and using
 the initial condition v = 200 at time t = 0, we arrived at the equati
on:" }}{PARA 0 "" 0 "" {TEXT -1 100 "                                 \+
                          v(t) = 20 * (29.62 * e ^ (-t/40) - 19.62)" }
}{PARA 0 "" 0 "" {TEXT -1 72 "To check on Maple we isolate the dv/dt b
y dividing through by 2 and get:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "v:=unapply(rhs(dsolve(\{diff(v(t),t)=-9.81-v(t)/40,v(
0)=200\},v(t))),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"vGR6#%\"tG6
\"6$%)operatorG%&arrowGF(,&$!++++CR!\"(\"\"\"-%$expG6#,$9$#!\"\"\"#S$
\"++++CfF/F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "This is the s
ame answer with the 20 distributed.  Next we ask Maple when the veloci
ty is zero." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(v(t)=0
,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+>9gZ;!\")" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 113 "To find the height at this point, we mus
t integrate the velocity function to get a function in terms of distan
ce." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "x:=unapply(\{int(v(t
),t)+c,x(0)=0\},t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xGR6#%\"tG6
\"6$%)operatorG%&arrowGF(<$,(9$$!++++CR!\"(-%$expG6#,$F.$!+++++D!#6$!&
'pB\"\"!%\"cG\"\"\"/-F$6#F;F;F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 79 "We know that at time t = 0, x(t) = 0, so c = 23696.  This match
es our solution." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "To complete t
he problem, we must find x when t = 16.48 sec." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 38 "x1:=t->23696-392.4*t-23696*exp(-t/40);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x1GR6#%\"tG6\"6$%)operatorG%&arrowG
F(,(\"&'pB\"\"\"9$$!%CR!\"\"-%$expG6#,$F/#F2\"#S!&'pBF(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x1(16.48);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#$\"*&>\"[`\"!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 213 "Sadly, in over an hour of Maple expirimentation, we could not \+
figure out how to get Maple to give us the x value for t = 16.48 sec. \+
 On a calculator, this simple plugging in of numbers gives    x(16.48)
 = 1534.8m." }{TEXT 256 62 "Dr. Sarah's comment - I put in x1(16.48) w
hich does the trick!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
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