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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "John Conowall and Sarah Th
ompson" }}{PARA 0 "" 0 "" {TEXT -1 16 "pg 96 problem #2" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 561 "Our given problem i
s to solve for the time that it takes a 400 lb. object to hit the grou
nd if it is dropped from a height of 500 ft.  We are not ignoring the \+
force due to air resistance.  This force is given by 10v(t).  Therefor
e, our d.e. to solve  is a simple sum of the forces problem: accelerat
ion = v' = g-10v(t)/m where g is the acceleration due to gravity, and \+
m is the mass of the object.  The acceleration due to gravity is 32 ft
/s^2.  To solve for time we have to differentiate v' and the different
iate our result to determine an equation for x(t).  " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "f:=(diff(v(t),t)=32-10*v(t)/400);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG/-%%diffG6$-%\"vG6#%\"tGF,,&\"#K
\"\"\"F)#!\"\"\"#S" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "dsolv
e(\{f,v(0)=0\},v(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#%\"t
G,&\"%!G\"\"\"\"-%$expG6#,$F'#!\"\"\"#S!%!G\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "Now tha
t we have solved for our velocity equation we can differentiate this t
o find our position equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 41 "h:=(diff(x(t),t)=1280-1280*exp(-1/40*t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"hG/-%%diffG6$-%\"xG6#%\"tGF,,&\"%!G\"\"\"\"-%$expG6
#,$F,#!\"\"\"#S!%!G\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ds
olve(\{h,x(0)=0\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%
\"tG,(F'\"%!G\"-%$expG6#,$F'#!\"\"\"#S\"&+7&!&+7&\"\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 129 "Now that we have our equation for posistion we can use x(t)=50
0 to solve for the time it takes for this object to hit the ground." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "evalf(1280*t+51200*exp(-1/40
*t)-51200=500);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,(%\"tG$\"%!G\"\"
\"!-%$expG6#,$F%$!+++++D!#6$\"&+7&F($!&+7&F(\"\"\"$\"$+&F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "solve(1280.*t+51200.*exp(-.25000000
00e-1*t)-51200. = 500.,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+F\"o
Ms&!\"*$!+o%QHY&F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 312 " Since the time for decent can't \+
have a negative value, we know that the positive value is the solution
 we are looking for.  We verified this result by plugging it back into
 our original x(t) equation. So now we have the time, in seconds, that
 it takes for the object to hit the ground from a height of 500 ft.  \+
" }}}}{MARK "12 0 0" 167 }{VIEWOPTS 1 1 0 1 1 1803 }