Mathematics HL IA 7

                                 

Modeling  the  trajectory  of  a  football  in  air   Mathematics  HL  exploration.  

                        Candidate  name:  Haitham  Wahid   Candidate  number:  001217-­‐0031   Online  course:  Pamoja  Education     Supervisor:  Ellen  Lawsky     Examination  season:  May  2016              

     

Table  of  contents   1.  INTRODUCTION:  ....................................................................................................  2   1.1  RATIONALE  ...........................................................................................................................  2   2.  DESCRIBING  THE  SITUATION  ............................................................................  2   3.  MODELING  THE  FLIGHT  OF  THE  BALL  ............................................................  3   3.1  FORCES  ACTING  ON  THE  BALL  ..............................................................................................  3   3.2  CONSTRUCTING  THE  TWO  DIFFERENTIAL  EQUATIONS.  .....................................................  4   3.3  SOLVING  FOR  THE  VELOCITIES  OF  THE  BALL  ......................................................................  5   3.3.1  Horizontal  velocity  ....................................................................................................  5   3.3.2  Vertical  velocity  ..........................................................................................................  6   3.4.  SOLVING  FOR  THE  DISPLACEMENTS  OF  THE  BALL  .............................................................  8   4.  THE  TRAJECTORY  CURVE  OF  THE  BALL  IN  AIR  ..........................................  10   5.  DETERMINATION  OF  THE  INITIAL  CONDITIONS  .......................................  11   5.1.  THEORETICAL  RESULT  .....................................................................................................  11   5.2.  EXPERIMENTAL  RESULT  ...................................................................................................  12   6.  CONCLUSION  AND  EVALUATION  OF  THE  RESULT  ......................................  13   6.1  CONCLUSION  ......................................................................................................................  13   6.2  EVALUATION  OF  THE  RESULTS  .........................................................................................  14   BIBLIOGRAPHY  ........................................................................................................  15   APPENDIX    ................................................................................................................  15    

     

   

 

   

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1.  Introduction:    

The   aim   of   this   exploration   is   to   find   a   mathematical   model   that   describes   the   trajectory   of   a   football   when   kicked   by   a   player   in   “free   kick”.   A   free   kick   is   a   kick   that   is   made   without   being   stopped   or   slowed   by   the   opponent;   hence   it   can   be   thought   of   as   a   golden   opportunity   for   a   team   to   score   a   goal.   However,   due   to   the   numerous   amount   of   factors   needed   to   be   taken   into   account   for   it   to   be   successful,   it   is   often   very   hard   to   make   the   correct   kick.   This   includes   the   kicking   power,   the   angle   of   the   kick,   the   spin   being  applied  to  the  ball  etc.   Football,  as  being  the  most  popular  sport  in  the  world,  is  becoming  an  increasingly   competitive  game.  As  a  result,  players  are  trying  to  find  different  tactics  to  improve  their   kicking  skills.  These  include  adding  a  spin  to  the  ball  or  kicking  it  with  a  very  high  speed   so   that   the   goalkeeper   cannot   predict   the   path   the   ball   would   take.   Understanding   the   mathematics   behind   a   football   kick   can   be   very   important   as   it   may   provide   some   necessary  conditions  to  enhance  players’  kicking  performance.    

1.1  Rationale      

My   motivation   for   choosing   this   topic   is   because   of   a   problem   I   am   used   to   face   frequently  in  my  football  training.  Most  often  when  I  kick  the  ball  freely  a  certain  distant   from   the   goalpost,   it   either   hit   the   defense   or   pass   way   over   the   goalpost.   Although   it   may  be  due  to  lack  of  practice,  it  is  sometimes  not  clear  to  me  the  angle  that  I  should  aim   to  kick  to  ball  at,  or  the  power  I  should  apply  to  it  (which  would  than  be  the  initial  speed   of   the   ball)   when   I   kick   it   from   a   certain   distant.   Is   it   possible   to   know   both   of   these   conditions  so  that  the  ball  enters  the  goalpost,  or  at  least  reach  that  point?  This  question   can   only   be   answered   if   I   can   find   an   equation   of   the   balls   trajectory   in   the   form   of   𝑦 = 𝑓(𝑥),   where  𝑦  is   the   height   reached   by   the   ball   and  𝑥  is   its   horizontal   displacement   from  the  point  where  it  was  kicked.  Obtaining  an  equation  of  motion  of  the  ball  would   give  us  very  useful  information  about  its  movement  in  air  and  therefore  provide  us  with   some  clues  on  how  to  kick  a  ball  successfully.  This  will  be  explored  in  this  project.    

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2.  Describing  the  situation    

𝒗𝟎   𝜃     Figure  1:  A  player  kicks  the  ball  with  an  initial  velocity  𝒗𝟎 and  angle  𝜃  to  the  horizontal.  The  diagram     is  not  to  scale.  

     

 

   

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  Let  us  suppose  that  a  player  kicks  the  ball  a  distant  of  𝑥  meters  from  the  goalpost   in   free   kick;   it   would   follow   a   curvature   path   as   shown   in   figure   1.   From   the   diagram   above   we   see   that   the   ball   has   to   reach   over   the   defense   while   at   the   same   time   pass   under   the   goalpost   if   it   were   to   be   successful.   Now   clearly,   this   is   not   the   only   way   to   score  a  goal;  professionals  use  a  variety  of  techniques  such  as  swerving  the  ball  around   the  defense  and  still  score  a  goal.  However,  in  this  exploration  it  will  be  assumed  that  the   ball   will   follow   a   path   similar   to   that   shown   above   to   reduce   the   problem   into   a   two-­‐ dimensional   motion.   Therefore,   to   proceed   further,   some   assumptions   will   be   stated.   These  are:     • The  drag  force  acting  on  the  ball  is  linearly  proportional  the  velocity  of  the  ball   but  opposite  in  direction.       • The  speed  of  the  wind  during  the  flight  of  the  ball  is  assumed  to  be  negligibly   small  to  exert  any  forces  on  the  ball.     • No  spin  is  applied  to  the  ball.   • The  ground  is  equally  flat  everywhere  in  the  field.  

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3.  Modeling  the  flight  of  the  ball    

3.1  Forces  acting  on  the  ball    

To   generalize   the   problem   let   us   assume   that   a   player   kicks   a   ball   of   mass   m (which   is   0.48kg   for   a   common   football)   with   an   initial   velocity v 0  (in   meters   per   second)  at  an  angle θ (in  degrees)  to  the  horizontal.  Generally,  as  the  ball  moves  through   air   multiple   forces   would   act   on   it,   and   including   all   these   forces   would   require   a   thorough  knowledge  of  the  aerodynamics.  Therefore,  for  the  purpose  of  simplicity,  I  will   be   assuming   that   no   forces   other   than   gravity   (𝑭𝑮 )  and   air   drag   𝑭𝑫    act   on   the   ball   during  its  motion.  The  force  of  gravity  acts  downwards  on  the  ball  and  has  a  magnitude   of  𝑭𝑮 = 𝑚𝒈,  where  𝑔  is  the  gravitational  field  strength1  and   is  9.81ms-­‐1.   In   addition   to   the   force   of   gravity,   air   drag   also   affects   motion   of   the   ball.   This   force   is   a   result   of   frequent   and   direct   collisions   between   the   air   particles   and   the   ball.   Using  newton’s  third  law  of  motion  we  can  deduce  that  the   direction   of   this   force   is   opposite   to   the   velocity   vector   of   the   ball   at   any   instant   in   time.   It   is   assumed   in   this   exploration   that   the   drag   force   is   linearly   proportional   to   the  velocity  of  the  ball  in  air,  thus:      

𝑭𝑫 ∝ 𝒗 ⟺ 𝑭𝑫 = 𝑏𝒗,  where  𝑏  is  a  constant  of   proportionality  and  has  a  value  of  0.47kgs-­‐1  for  the   particular  shape  of  the  football,  according  to  an  article   published  by  Takeshi    Asai  and  Kazuya  Seo2.                                                                                                                     1  IB  physics  data  booklet  for  first  examination  2016   http://www.springerplus.com/content/2/1/171  

 

   

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Figure  2:  A  free  body  diagram  repressing   the  forces  acting  on  the  ball  and  the   velocity  vector  at  an  instant  in  time.  The   drag  force  is  opposite  in  direction  to  the   velocity  of  the  ball  

We  can  now  display  the  two  forces  on  a  free  body  diagram,  as  shown  in  the  figure   2.   The   dotted   arrow   is   the   velocity   vector   at   an   instant   point   in   time   directed   with   an   angle  𝛼  to   the   horizontal.   Using   the   vertically   opposite   angles   theorem,   we   can   deduce   that  𝛼  is  also  the  angle  between  the  drag  force  and  the  horizontal  𝑥  axes.      

3.2  Constructing  the  two  differential  equations.     In  order  to  determine  the  equation  of  the  trajectory  of  the  ball  in  air,  we  need  to   consider   the   actual   motion   as   a   superposition   of   two   separate,   independent   motions-­‐   horizontal  𝑥  and   vertical   y.   Since   neither   direction   affect   the   other   we   add   all   forces   that   act  on  each  separately.  To  do  that  we  incorporate  newton  second  law  of  motion  to  our   discussion,  which  dictates  that    “the  sum  of  forces  on  a  body  is  equal  to  the  mass  times   its   acceleration3”   Given   the   acceleration   is,   by   definition,   the   derivative   of   velocity   and   the   second   derivative   of   displacement   with   respect   to   time,   we   obtain   two   differential   equations  for  the  ball’s  trajectory.  These  two  differential  equations  will  then  be  solved  to   give  the  two  solutions  𝑥 𝑡  𝑎𝑛𝑑  𝑦(𝑡)  for  both  the  horizontal  and  vertical  displacements   respectively.   Combining   the   equations   into   one   of   the   form  𝑦 = 𝑓(𝑥)  would   therefore   give  us  the  actual  trajectory  of  the  ball.     Thus,  we  start  first  by  adding  all  forces  that  act  on  each  direction  of  motion.  For  the   horizontal  direction  of  motion:         dv ∑ Fx = m dtx = −bv cos(α )     The  minus  sign  here  only  accounts  for  the  direction  of  the  drag  force  component  as   being  opposite  to  the  velocity  of  the  ball  in  the  𝑥  direction.  Notice  that  we  can  rewrite   the  expression  above  in  terms  of  the  horizontal  velocity,  as vx = v cos(α ) ,  thus     dv m x = −bvx dt Similarly,  the  sum  of  forces  in  the  vertical  direction:       may = ∑ Fy = −bvsin(α ) − mg   But   vy = vsin(α )   Thus:     dvy = −bvy − mg dt   Again,   we   assign   the   vertical   component   of   the   drag   force   to   be   negative   as   it   is   always   in   the   opposite   direction   of   the   velocity   vector   in   the   y   direction.     As   for   the   gravitational  force,  it  is  always  directed  towards  the  negative  y  direction,  hence  it  would   be  negative.   We   thus   obtain   two   ordinary   first   order   differential   equations   for   the   horizontal   and  vertical  motion.  To  solve  for  the  particular  solutions  of  these  differential  equations     we   are   required   to   introduce   some   initial   conditions.   Since   the   ball   is   kicked   by   the   player   with   an   initial   velocity   of v0 at   an   angle   of   θ to   the   horizontal,   then   at   time t = 0 we   have:                                                                                                                     m

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 K.  A.  Tsokus;  Physics  for  the  IB  dipmola  2014  

   

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vx (0) = v0 cos(θ ) And vy (0) = v0 sin(θ )

 

3.3  Solving  for  the  velocities  of  the  ball     3.3.1  Horizontal  velocity    

Let   us   now   solve   the   differential   equations   developed   earlier   for   the   velocities   of   the  ball  in  the  horizontal  and  vertical  directions.  Starting  with  the  horizontal  velocity  we   notice   that   it   is   a   first   order   equation   that   can   be   solved   using   the   method   variable   separation,  i.e.     dvx = −bvx dt   1 b ⇒ dvx = − dt vx m m

Integrating  both  sides  gives:  

∫

dvx b = − dt vx ∫ m

 

b ln(vx (t)) = − t + C m

But : v x (0) = v0 cos(θ )

⇒ ln ( v0 cos(θ )) = C thus b t m   Using  logarithmic  rules  and  taking  the  exponent  of  both  sides  gives:   ln(vx (t)) − ln(v0 cos(θ )) = −

  b − t ⎛ v (t) ⎞ b vx (t) m ln ⎜ x = − t ⇔ = e m v0 cos(θ ) ⎝ v0 cos(θ ) ⎟⎠

⇒ vx (t) = v0 cos(θ )e

b − t m

 

  We   therefore   see   that   the   horizontal   component   of   the   velocity   takes   an   exponential   decay   function,   that   it,   it   decreases   exponentially   with   time.   It   is   quite   a   realistic   result,   as   the   ball   does   not   travel   forever   when   kicked   with   some   initial   horizontal  velocity.  It  always  reaches  a   stop  at  on  point  in  time  due  to  air  resistance  and   other  forces  opposing  the  motion.  This  is  clearly  seen  from  the  function  by  realizing  that   as t → ∞,vx → 0 .   This   does   not,   however,   mean   that   it   would   actually   take   an   infinite   amount  of  time  for  ball  to  stop  progressing  in  the  𝑥  direction,  as  in  most  cases  in  real  life   the  time  can  be  short.      

 

   

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Lets   take   a   look   at   the   affect   of   air   drag   in   more   detail.   The   graph   to   the   left   shows   two   exponential   curves   for   the   velocity   of   the   ball   as   function   of   time   that   have   been   drawn   using   Desmos  calculator.      Both   curves   have   the   same   initial   conditions   but   with   different   drag   coefficients.     The   blue   gave   has  b   value   of   0.47kgs-­‐1   but   the   red   curve   was   drawn   with   twice   the   drag   coefficient  as  that  of  the  blue  one.    The  scales  of   the  axes  are  arbitrary.   It   is   shown   that   large   drag   coefficients   result   in   a   large   deceleration   and   thus   increases   the   rate   of   decay.   If   the   horizontal   velocity   approaches   zero   before   the   ball   has   reached   the   ground   again   when   kicked,   it   Figure  3:  Horizontal  velocity  as  a  function  of  time  for   would   not   progress   further   in   the  𝑥  direction   two  different  drag  coefficients   and   thus   appears   to   be   in   free   fall.   This   is   a   commonly   seen   when   a   badminton   ball   is   shot   in   air.   Since   it   has   a   smaller   mass   its   horizontal   velocity   decrease   more   rapidly   a   football   and   hence   would   be   in   free   fall   a   certain  amount  of  time  after  it  was  shot.     3.3.2  Vertical  velocity    

Now  solving  for  the  vertical  velocity  of  the  ball  when  it’s  in  air  from  the  differential   equation  of  the  vertical  motion,  i.e.     dv                                                                                                                                                   m y = −bvy − mg   dt   ! By   adding  ! 𝑣!  to   both   sides   of   the   equation   we   obtain   a   first   order   exact   differential   equation:   dvy b + vy = −g   dt m Solving   this   differential   equation   requires   finding   its   integrating   factor I(t) and   than   multiplying  it  with  both  sides  of  the  equation.  Thus:     b b dt t I(t) = e ∫ m = e m   Multiplying  𝐼 𝑡  with  both  sides  of  the  equation:     b b b t dvy t t b m m m e + vy e = −ge   dt m Note  here  that  the  product  rule  of  implicit  differentiation  has  been  applied  to  LHS.   Mathematically:   b b b t dv t t⎞ b d⎛ e m y + vy e m = ⎜ vy ⋅ e m ⎟ dt m dt ⎝ ⎠  

 

   

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  Thus  we  have  inversed  the  product  rule.  Now  integrating  both  sides  of  the  equation  with   respect  to  𝑡  to  undo  the  product  rule  on  the  left  hand  side:       b t d ⎛ mb t ⎞ m v e dt = − ge ∫ dt ⎜⎝ y ⎟⎠ ∫ dt   b b t t m ⇒ vy e m = − ge m + C b Since vy (0) = v0 sin(θ )   Thus : mg mg v0 sin(θ ) = − + C ⇒ C = v0 sin(θ ) + b b   b b t mg m t mg ⇒ vy e m = − e + v0 sin(θ ) + b b     Solving for vy (t) and simplifying we obtain:     mg ⎞ − mb t mg ⎛ ∴vy (t) = ⎜ v0 sin(θ ) + ⎟e − ⎝ b ⎠ b     This  is  another  exponential  decay  function.  However  unlike  for  the  horizontal   velocity  of  the  ball  that  approaches  zero  with  an  increase  in  time,  the  vertical  velocity   approaches  a  constant  value.  Taking  the  limit  of  the  function:       ⎛⎛ mg ⎞ − m t mg ⎞ mg ⎞ − m t mg ⎛ lim ⎜ ⎜ v0 sin(θ ) + e − = lim ⎜ v0 sin(θ ) + ⎟ ⎟e − ⎟ t→∞ ⎝ ⎝ b ⎠ b ⎠ t→∞ ⎝ b ⎠ b b

⇒−

b

mg b

  Therefore,   as   time   increases   the   vertical   velocity   approaches   a   constant   velocity   of   !" − !  and   in   the   downward   direction.   This   is   often   known   as   the   terminal   velocity   and   is   reached   when   the   net   force   on   the   body   in   the   vertical   direction   is   zero.   This   constant   is   represented  by  the  horizontal  asymptotes  by  the  function  𝑣! (𝑡).     Using   Desmos   calculator   I   have   drawn   two   curves   for   the   vertical   velocity   of   the   ball.   Both   curves   have   the   same   initial  conditions  and  the  scales  of  the  axis   are   arbitrary.   Again,   looking   at   the   effect   of   air   drag   on   the   motion   of   the   ball   we   notice   that   for   large   value   of  𝑏  the   vertical   velocity   decreases   faster   and   hence   the   height   reached   by   the   ball   would   be   minimized.   This   is   seen   from   the   areas   under   the   two   curves   and   the   positive   axis.          

   

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Figure  4:  The  vertical  velocities  as  a  function  of  time  for   two  different  drag  coefficients  

 

3.4.  Solving  for  the  displacements  of  the  ball    

Having  found  the  velocity  functions  of  the  ball  in  air  we  now  move  on  to  the  next   step   of   determining   the   displacement   functions.   This   is   done   the   realizing   that   the   velocity   is   the   derivative   of   displacement   with   respect   to   time.   Thus   for   the   horizontal   displacement:     x(t) = ∫ vx (t)dt ⇒ ∫ v0 cos(θ )e

b − t m

dt = −

b   − t m v0 cos(θ )e m + C b

Since  the  initial  horizontal  displacement  of  the  ball  is  zero  when  it  was  kicked:    

m v0 cos(θ ) + C = 0 b   m ⇒ C = v0 cos(θ ) b x(0) = −

Now:    

b − t m m v0 cos(θ ) − v0 cos(θ )e m b b     Simplifying  further  we  obtain:   b − t⎞ ⎛ m ∴ x(t) = v0 cos(θ ) ⎜ 1− e m ⎟   b ⎝ ⎠

x(t) =

 

Similarly,  for  the  vertical  displacement:  

  y(t) = ∫ vy (t)dt = −

m⎛ m ⎞ − mb t m ⎜⎝ v0 sin(θ ) + g ⎟⎠ e − gt + C b b b  

 

One  of  the  assumptions  made  earlier  was  that  the  field  is  flat  everywhere.  We  can   use  this  assumption  to  conclude  that  the  initial  height  relative  to  any  point  in  the  field  is   zero  when  the  ball  was  kicked.  Therefore:     m⎛ m ⎞ y(0) = 0 ⇒ ⎜ v0 sin(θ ) + g ⎟ = C   b⎝ b ⎠   m⎛ m ⎞ m⎛ m ⎞ bt m ⇒ y(t) = ⎜ v0 sin(θ ) + g ⎟ − ⎜ v0 sin(θ ) + g ⎟ e− m − gt b⎝ b ⎠ b⎝ b ⎠ b     Simplifying  the  expression  further  reveals  that:  

  b t m⎛ m ⎞⎛ m −m ⎞ y(t) = ⎜ v0 sin(θ ) + g ⎟ ⎜ 1− e ⎟ − gt ⎝ ⎠ b b ⎝ ⎠ b  

 

   

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We  thus  obtain  two  parametric  equations  for  the  path  for  the  ball  in  terms  of  time,   i.e.       x(t) =

b − t⎞ ⎛ m v0 cos(θ ) ⎜ 1− e m ⎟   b ⎝ ⎠

⇒y=

m⎛ m ⎞ m⎛ m ⎞ − mb t m v sin( θ ) + g − v sin( θ ) + g ⎟ e − gt   ⎜ 0 ⎟ ⎜ 0 b⎝ b ⎠ b⎝ b ⎠ b

 

             (1)                    (2)  

 

Looking   at   the   two   equations   above   we   may   notice   that   both   have   time  𝑡  as   a   common   variable.     We   can   solve   for   t   using   equation   (1)   and   then   substituting   it   into   equation   (2)   to   acquire   a   well   define   relation   between   the   horizontal   and   vertical   components  of  the  motion.         Now:   b b − t⎞ − t⎞ ⎞ ⎛ ⎛ m b⎛ x m m                                 x = v0 cos(θ ) ⎜ 1− e ⎟ ⇒ ⎜ 1− e ⎟ = ⎜                  (3)   b ⎝ ⎠ ⎝ ⎠ m ⎝ v0 cos(θ ) ⎟⎠   Solving  for  𝑡:     b − t ⎞ ⎞⎞ b⎛ x m ⎛ b⎛ x m                        (4)   1− ⎜ = e ⇒ t = − ln 1− m ⎝ v0 cos(θ ) ⎟⎠ b ⎜⎝ m ⎜⎝ v0 cos(θ ) ⎟⎠ ⎟⎠

  Substituting  (3)  and  (4)  into  (2)     y=

⎞ m ⎛m ⎛ ⎞⎞⎞ m⎛ m ⎞ b⎛ x b⎛ x   + g ln 1− ⎜⎝ v0 sin(θ ) + g ⎟⎠ ⋅ ⎜ ⎜ b b m ⎝ v0 cos(θ ) ⎟⎠ b ⎝ b ⎜⎝ m ⎜⎝ v0 cos(θ ) ⎟⎠ ⎟⎠ ⎟⎠

 

Simplifying  further:    

2 ⎞ ⎛ m⎞ ⎛ ⎛ ⎞⎞⎞ m ⎞⎛ x b⎛ x ⎛ ⇒ ⎜ v0 sin(θ ) + g ⎟ ⎜ + g ln 1− ⎜ ⎟ ⎝ b ⎠ ⎝ v0 cos(θ ) ⎟⎠ ⎝ b ⎠ ⎜⎝ ⎜⎝ m ⎜⎝ v0 cos(θ ) ⎟⎠ ⎟⎠ ⎟⎠     Thus,   the   equation   of   motion   of   a   ball   kicked  with   an   initial   velocity   v 0  at   an   angle   θ to  the  horizontal  and  being  effected  by  gravitational  force  and  air  resistance  is  defined   as:  

2 ⎛ ⎞ ⎛ m⎞ ⎞⎞ mg ⎞ ⎛ x b⎛ x ⎛ y(x) = ⎜ v0 sin(θ ) + + ⎜ ⎟ g ln ⎜ 1− ⎜ ⎟ ⎜ ⎟ ⎝ b ⎠ ⎝ v0 cos(θ ) ⎠ ⎝ b ⎠ ⎝ m ⎝ v0 cos(θ ) ⎟⎠ ⎟⎠

(5)    

θ ∈]0,90[, x ∈[0, R] and y ∈[0,hmax ]

  Where  𝑅  is  the  range  of  the  shot  and  ℎ!"#  is  the  maximum  height  traveled  by  the   ball/m.    

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9  

4.  The  trajectory  curve  of  the  ball  in  air        To   visualize   the   shape   of   the   trajectory   of   equation   (5)   that   we   have   developed   earlier,   I   have   drawn   it   for   different   values   of   v0 and θ using   Desmos   calculator   and   noticed   that   both   these   parameters   change   the   trajectory   of   the   ball   in   a   very   unique   way.   Figure   5   depicts   three   trajectories   of   a   kick   with   different   initial   conditions   displayed   in   table   1.   The   black   and   red   balls   were   kicked   at   the   same   angle  𝜃  but   with   different   initial   velocities  𝑣! .   We   notice   that   the   red   ball   covers   a   longer   distant  𝑥  and   reaches  a  higher  point  y  because  it  has  a  larger  initial  velocity  in  both  directions.           Color   Initial   Initial     angle/  °   speed/ms-­‐1     Red   25   25   Table  1:  Initial  values.       Green   35   20     Black   25   20    

  Figure  5:  The  theoretical  trajectory  of  three  balls  kick  with  different  initial  conditions  

  Now  comparing  the  green  ball  with  the  black  ball,  they  both  have  been  kicked  with   the  same  initial  velocities  𝑣!  but  different  angles   θ .  The  green  ball  is  kicked  at  a  larger   angle  and  we  observe  that,  though  it  has  a  smaller  range  compared  with  the  black  ball,  it   reaches   a   higher   point   in   the   vertical   direction.   This   is   because   increasing   the   angle   increases  the  initial  velocity  in  the  vertical  direction  and  as  a  result,  increases  the  height   of   the   ball;   while   at   the   same   time   reduces   the   initial   velocity   in   the   horizontal   direction   and  therefore  minimize  the  range  of  the  kick.      

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5.  Determination  of  the  initial  conditions        

5.1.  Theoretical  result    

Let   us   now   go   back   to   our   problem   of   determining   at   what   angle   and   with   what   initial  velocity  should  a  football  player  kick  the  ball  in  free  kick  when  he  is  at  a  certain   distant  𝑥  from   the   goalpost.   This   distance   can   be   chose   between   25-­‐30m   but   I   will   choose   20m   because   I   usually   perform   the   kick   for   that   distant.   After   doing   some   research   about   the   distant   from   the   kicker   and   the   wall   (defense),   I   found   that   it   is   approximately   9   meters   and   the   wall   can   have   a   maximum   height   of   about   2.2m   depending  on  how  tall  the  players  are.  Having  found  these  values,  I  went  to  the  nearest   football  field  in  the  vicinity  and  measured  the  height  of  the  goalpost  and  found  it  to  be   around   2.5m.   Thus   for   the   ball   to   enter   the   goalpost   any   value   of   the   interval   [0𝑚, 2.5𝑚[  can   be   chosen.   2.3m   is   a   good   choice   because   it   would   be   a   bit   harder   for   the   goalkeeper   to   catch   as   its   quite   high.   We   have   now   all   the   necessary   values   for   calculating   v0  and   θ .    

𝒗𝟎   𝒗𝟎   𝜃  𝜃  

Figure  6:  A  possible  trajectory  of  a  football  kick  with  an  initial  velocity  𝒗𝟎  and  an  angle  𝜃  that  would  successfully   enter  the  goalpost.    

 

Let  us  restate  the  parameters  in  our  problem  that  we  have  mentioned  previously.   𝑚 = 0.48𝑘𝑔   𝑏 = 0.47𝑘𝑔𝑠 !!   𝑔 = 9.81𝑚𝑠 !!     Substituting   these   values   into   equation   (5),   we   obtain   two   simultaneous   equations   in   terms   of   v0 and   θ  that   are   implicit   in   nature,   i.e.   we   cannot   basically   express   one   unknown  in  terms  of  the  other.  Now,  from  figure  6  we  see  that  when   x = 9m ⇒ y = 2.2m and  when   x = 20m ⇒ y = 2.5m .  It  follows  that   y(9) = 2.2m  and   y(20) = 2.5m .  Hence:       2 ⎛ 0.47 ⎛ ⎞ ⎛ 0.48 ⎞ ⎞⎞ 4.7 ⎞ ⎛ 9 9 ⎛ 2.2 = ⎜ v0 sin(θ ) + +⎜ ⎟⎠ ⎜ ⎟⎠ 9.81ln ⎜ 1− ⎟ ⎜ ⎝ ⎝ 0.47 ⎝ v0 cos(θ ) ⎠ 0.47 ⎝ 0.48 ⎝ v0 cos(θ ) ⎟⎠ ⎟⎠   2 ⎛ 0.47 ⎛ 20 ⎞ ⎞ 4.7 ⎞ ⎛ 20 ⎞ ⎛ 0.48 ⎞ ⎛ 2.5 = ⎜ v0 sin(θ ) + +⎜ ⎟⎠ ⎜ ⎟⎠ 9.81ln ⎜ 1− ⎟ ⎝ ⎝ 0.47 ⎝ v0 cos(θ ) ⎠ 0.47 ⎝ 0.48 ⎜⎝ v0 cos(θ ) ⎟⎠ ⎟⎠

 

   

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  Using  graphical  display  calculator  to  find   the  point  of  intersection  we  obtain  the  following   values:     𝑣! ≈ 30.9𝑚𝑠 !!   𝜃 ≈ 18.3°     These  are  the  required  values  for  the  initial   velocity  and  the  angle  at  which  the  ball  should  be   kicked  at  so  that  it  would  have  enough  height  to   pass   over   the   defense   (wall)   while   at   the   same   time   reaches   just   under   the   top   of   the   goalpost,   as  shown  by  the  trajectory  of  the  ball  in  figure  9.          

(30.94,  18.32)  

Figure  7:  A  GDC  was  used  to  determine  𝑣!  and  𝜃.    

Figure  8:  The  final  (theoretical)  trajectory  of  the  ball  that  is  regarded  to  successfully  enter  the  goalpost.    

   

5.2.  Experimental  result    

Since  there  are  no  literature  values  that   can  be  compared  with  the  result  obtained,   the  only  way  left  of  determining  the  accuracy  of  the  result  is  by  testing  it  in  practice.  This   means  that  I  will  try  to  kick  the  ball  with  the  same  initial  velocity  and  angle  as  that  found   in  this  exploration.     There  are  of  course  many  ways  of  experimenting  the  values  but  the  most  efficient   one  in  (my  opinions)  is  by  using  a  smart  ball.  This  ball  is  an  extraordinary  because  the   interesting  thing  about  it  is  that  it  contains  a  small  device  located  at  its  center  that  has   features   such   detecting   the   initial   velocity   applied   to   the   ball,   the   spin   if   any,   the   trajectory   of   the   ball   and   many   others.   Unfortunately   though,   it   does   not   record   the   kicking   angle.   Therefore   I   will   be   using   an   instrument   found   in   our   physics   lab   called   Winkeltronic  and  set  it  on  angle  of  18.3  and  place  it  at  a  side  so  that  I  know  roughly  how   high  I  should  kick  the  ball.  Once  I’ve  kicked  the  ball  the  device  would  record  the  data  of   the  kick  and  send  it  via  Bluetooth  to  an  app  on  my  smartphone  called  Adidas  Smart  Ball   and  the  data  would  be  displayed  on  my  phone.    

 

   

12  

  Since  it  was  a  free  kick  I  had  to  ask  my  friend  to  stand  in   the   defense   for   the   purpose   of   blocking   the   ball   but   did   not   ask   for   a   goalkeeper   as   the   purpose   of   this   whole   exploration   is  for  the  ball  to  reach  the  goalpost,  not  necessarily  entering   it   if   there   was   a   goalkeeper.   After   hours   of   kicking   I   finally   performed   one   that   entered   the   goalpost   and   that   gave   an   incredible   result   that   resembles   the   theoretical   one   (see   figure  9).   The  experimental  result  is  shown  in  the  snapshot  in  the   figure  to  the  right.    The  kick  was  performed  a  distant  of  20m   from  the  goalpost  and  9m  from  the  defense.  The  angle  of  shot   was   estimated   to   be   20°   with   an   uncertainty   of   ±4°   and   the   velocity,   as   seen   from   the   snapshot,   is   64mph   (≈   29ms-­‐1).   Comparing   experimental   result   with   the   theoretical   one   obtained   earlier   we   see   that   they   are   quite   close   to   each   other,   which   is   very   pleasing   to   see!   For   other   results   the   reader  may  refer  to  the  appendix.     Figure  9:  a  snapshot  of  the    

_______________________________________________    

experimental  data  obtained  from   Adidas  smart  ball  app  

6.  Conclusion  and  Evaluation  of  the  result     6.1  Conclusion       The  aim  of  this  exploration  was  the  study  the  motion  of  the  ball  when  kicked  by  a   player   in   air.   The   equation   of   the   trajectory   was   successfully   derived   by   listing   a   few   assumptions  that  had  simplified  the  problem  into  a  two  dimensional  motion.  As  a  result,   two   differential   equations   were   constructed   for   the   motion   of   the   ball   in   the   x   and   y   direction  which  were  solved  to  give  the  actual  trajectory.  Furthermore  the  affect  of  air   drag  on  the  on  the  trajectory  was  also  studied  in  detain  and  was  concluded  that  as  the   drag  coefficient  increases  the  overall  displacement  of  the  ball  decreases  in  both  x  and  y   directions.     In   addiction   to   the   study   of   the   motion   of   the   ball,   the   intension   was   also   to   determine  the  initial  conditions  necessary  of  it  to  enter  the  goalpost  when  kicked  by  a   player  in  free  kick.  Both  the  initial  velocity  𝑣!  and  the  angle  𝜃  of  the  kick  were  calculated   from   the   equation   of   motion   and   were   than   compared   with   an   experimental   result.   It   was   shown   that   the   theoretical   result   was   in   good   agreement   with   that   of   the   experimental  and  therefore  completes  the  aim  of  this  exploration.                    

 

   

13  

 

6.2  Evaluation  of  the  results    

Although   we   showed   that   the   theoretical   result   was   similar   to   the   experiential   result,   the   difference   is   not   negligible.   This   is   because   the   list   of   assumptions   that   we   have   made   has   reduced   the   reliability   and   credibility   of   the   result.   For   example,   the   speed   of   the   wind   was   assumed   to   be   zero   throughout   the   flight   of   the   ball,   while   in   reality  there  was  a  constant  flow  of  air,  which  I  felt  when  I  kicked  the  ball.  This  would   have   the   affect   of   altering   the   path   of   the   ball   and   thus   would   change   the   initial   conditions.     Furtherer,  It  was  assumed  that  no  spin  was  applied  on  the  ball.  Although  this  can   be  a  valid  assumption  If  the  player  does  not  apply  any  spin  what  so  ever  on  the  ball,  in   my  case  I  tried  to  do  my  best  to  avoid  spinning  the  ball  but  couldn’t.  It  felt  as  If  the  ball   adds  spin  on  itself,  which  actually  turned  out  to  be  the  case.  After  doing  some  research,  I   found  that  the  movement  of  the  wind  is  what  causes  the  ball  to  spin.  So  even  if  you  don’t   spin  it,  the  wind  would  force  it  to  spin.  But  what  does  spinning  the  ball  have  to  do  with   its   trajectory?   Well,   here   is   where   thing   could   get   very   complicated.   A   simple,   but   incomplete   explanation   is   that   the   spin   applied   to   the   ball   would   generate   a   force   that   acts  normal  to  the  motion  and  spinning  axis,  which  again,  would  alter  the  path  traveled   by  the  ball.     Therefore,   the   model   derived   for   the   motion   of   the   ball   has   many   limitations.   However,   it   is   not   completely   unrealistic   as   it   takes   into   consideration   an   important   factor,   which   is   the   air   resistance   that   is   at   the   very   heart   of   aerodynamics,   and   thus   makes  it  a  practical  model  that  can  be  applied  to  real  world.     In   addition   to   the   example   made   about   determining  𝑣! and  𝜃 ,   one   can   even   calculate   the   optimum   angle  𝜃  to   maximize   the   range   of   the   kick   for   a   give   constant𝑣! .   It   is   very   essential   to   all   goalkeepers   to   kick   the   ball   with   the   maximum   possible   range   for   two  main  reasons;  (1)  is  to  save  the  goal  area  by  shooting  the  ball  very  far  and  that  (2)   would  create  more  opportunities  for  the  team  to  score  a  goal  as  they  would  receive  the   ball  in  a  region  close  to  the  opponents  goalpost.     It   is   interesting   to   note   that,   although   the   equation   of   motion   was   derived   from   modeling   the   trajectory   of   a   football,   it   is   not   only   limited   to   this   sport.   It   can   work   equally  well,  if  not  better,  in  other  sports  such  as  golf  or  baseball  as  the  aerodynamics   and,   of   course,   the   mathematics   associated   with   the   motion   of   spherical   bodies   is   the   very   similar.   By   altering   the   parameters   of   the   model   (equation   3),   it   is   possible   to   describe  different  situations  in  real  life.     Finally,   exploring   the   motion   of   a   football   was   a   very   fascinating   and,   of   course,   rewarding   topic.   This   is   because   it   gave   me   the   opportunity   to   apply   my   calculus   knowledge   to   model   a   real   life   situation   and,   even   more,   apply   it   to   my   favorite   sport   (Football).   Calculus   and   in   particular   the   concept   differential   equations  has   always   been   by  far  my  favorite  area  of  mathematics,  but  seeing  that  I  can  use  it  to  improve  my  self  in   football   makes   me   shake   with   excitement   in   exploring   the   depth   of   its   usefulness.   Therefore   I   have   decided   to   investigate   this   exploration   further   in   the   future   by   analyzing   the   affect   of   applying   a   spin   to   the   ball   on   its   trajectory   after   I   have   had   a   complete  understanding  the  physics  of  aerodynamics.  Until  then,  I  wouldn’t  consider  my   work  to  be  complete.            

 

   

14  

Bibliography     K.  A.  Tsokos:  Physics  for  the  IB  diploma  Sixth  edition  (2014)  Cambridge  press.       Marcelo  Alonso  &  Edward  J.  Finn  Physics,  Curvature  motion.       Modeling  vibration  of  airplane  wing  using  differential  equation     Vioh  2014.     IB  physics  data  booklet  for  first  examination  2016     http://www.thefa.com/football-­‐rules-­‐governance/laws/football-­‐11-­‐11/law-­‐ 13-­‐-­‐-­‐free-­‐kicks     http://guide.alibaba.com/shop/sport-­‐silhouette-­‐basketball-­‐player-­‐dribbling-­‐ car-­‐tablet-­‐vinyl-­‐deca_9560907.html     http://www.springerplus.com/content/2/1/171     Last  visited  2013.     Note:       All  diagrams  have  been  personally  drawn  by  the  author/  Haitham  Wahid  except  for  the   figures  of  the  players,  the  ball  and  of  course  the  goalpost  in  figure  (6).         Appendix.       Here  are  just  a  few  results  (out  of  many)  that  I  have  collected  while  trying  to  make  my   Ideal  kick.                                  

 

   

15