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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 8 "2280 - 2" }}{PARA 257 "" 
0 "" {TEXT -1 17 "Friday January 26" }}{PARA 258 "" 0 "" {TEXT -1 17 "
Computation sheet" }}{PARA 259 "" 0 "" {TEXT -1 22 "Example 2, section
 2.3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "y:= t-> -245*t + 294*25*(
1 - exp(-.04*t));\n   #height y(t)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%\"yGR6#%\"tG6\"6$%)operatorG%&arrowGF(,(9$!$X#\"%]t\"\"\"*&F/F0-%$ex
pG6#,$F-$!\"%!\"#F0!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 50 "v:= t -> 294*exp(-.04*t) - 245;\n   #velocity v(t) " }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"vGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%
$expG6#,$9$$!\"%!\"#\"$%H\"$X#!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 64 "25*ln(294.0/245);\n   #by hand we can set v(t)=0 an
d solve for t:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+?*Q!eX!\"*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "solve(v(t)=0,t);\n   #or we \+
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\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "y(4.558038920); \n  \+
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "solve(y(t)=0,t);  \n   #find
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!\"*$\"\"!F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "9.410949931
 - 4.558038920; \n   #time descending" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$\"+65\"H&[!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "v(9.4
10949931); \n   #speed when it lands " }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$!*$4tAV!\"(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 213 "Conclusions:  bolt rises for 4.56 seconds, to a height o
f 108.3 meters.  Then it spends 4.85 seconds descending,  landing with
 a velocity of -43.3 meters per second.  We can see these facts graphi
cally if we plot:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
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Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 
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