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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Bouncing Ball by David Blo
od" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart: with(plots):
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 439 "The following problems are a
ll related in the sense that each part is a representation of a bounci
ng ball with a coefficient of restitution equal to .8. The coefficient
 of restitution for the ball used for this project was found by a seri
es of drops and eyeball calculations of the height obtained on a meter
 stick. The average height of the rebound was found and used to determ
ine the initial rebound velocity verses the impact velocity. " }}
{PARA 0 "" 0 "" {TEXT -1 169 "coefficient of restitution = (rebound ve
locity)/(impact velocity) The method of finding these velocities was d
ependent of conservation of energy (ie E-initial = E-final)" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 167 "The following \+
equation represents the height at any given moment of a falling object
 where acceleration is -9.8m/s^2, initial height is y0, and initial ve
locity is v0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "eq1:=y0+v0
*t-4.9*t^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G,(%#y0G\"\"\"*&%
#v0GF'%\"tGF'F'*$)F*\"\"#\"\"\"$!#\\!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 78 "The initial height will be 150 meters above the ground. (
y0=0 is ground level)" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "y0:=150;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#y0G\"$]\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "In
itial velocity is zero (v0=0)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 6 "v0:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v0G\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 61 "For the initial fall the equation for hei
ght is the following" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "eq1;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"$]\"\"\"\"*$)%\"tG\"\"#\"\"\"$
!#\\!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Time when ball hits \+
ground is found when eq1 is zero" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "t1:=fsolve(eq1(t)=0,t=0..10);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#t1G$\"+_L$G`&!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 52 "plot from start till ball hits ground the first time" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "a:=plot(eq1,t=0..t1,color=bl
ue):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 147 "First rebound velocity---Velocity with which ball leaves groun
d the first time is:   (.8 * (velocity with which ball struck ground)=
.8*(v0-4.9*t1))" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Vr1:=.8*
9.8*(t1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Vr1G$\"+[8uPV!\")" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "equation for second flight with ne
w initial velocity (Vn1)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 
"eq2:=Vr1*(t-t1)-4.9*(t-t1)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e
q2G,(%\"tG$\"+[8uPV!\")$!+++++C!\"(\"\"\"*$),&F&F-$!+_L$G`&!\"*F-\"\"#
\"\"\"$!#\\!\"\"" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 44 "time at which ball strikes ground again (
t2)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "t2:=fsolve(eq2(t)=0,
t=6..20);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t2G$\"+sm`Q9!\")" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "plot from time ball leaves ground
 till it strikes again (t=t1..t2)--ie first impact to second impact" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "b:=plot(eq2,t=t1..t2,color
=red):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 93 "new rebound velocity (Vn2) will be .8 of the former rebou
nd velocity (conservation of energy)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "Vn2:=.8*Vn1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Vn
2G$\"+yI>qM!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "equation for p
ath of third flight" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "eq3:
=Vn2*(t-t2)-4.9*(t-t2)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G,(
%\"tG$\"+yI>qM!\")$!+-++#*\\!\"(\"\"\"*$),&F&F-$!+sm`Q9F)F-\"\"#\"\"\"
$!#\\!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "time when third fli
ght touches down" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "t3:=fso
lve(eq3(t)=0,t=15..40);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t3G$\"+T
$Rn9#!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "plot third flight fr
om t=t2..t3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "c:=plot(eq3,
t=t2..t3,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "d
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0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "This represents the f
irst two bounces of a ball with a coefficient of restitution equal to \+
8/10" }}{PARA 0 "" 0 "" {TEXT -1 145 "Theoretically, if each subsequen
t bounce was perfect (ie. perfect vacuum, perfect surfaces, etc.) then
 the ball would continue to bounce for ever" }}{PARA 0 "" 0 "" {TEXT 
-1 64 "However, as t->infinity the rebound velocity would approach zer
o" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "For \+
example, on the fiftieth rebound the velocity would be .0007738789164 \+
m/s" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "9.8*t1*.8^50;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+k\"*yQx!#8" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 156 "To see a life like demonstration of a bouncing ball
, you can run the java script available at http://www.phy.ntnu.edu.tw/
java/bouncingBall/bouncingBall.html" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
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