Perfect Free Kick

Words: 1916
Pages: 8

How to take the perfect free kick

Mathematics Level I
2014/15

Amber Lynch
11/05/2015

Abstract
My report is looking into how to take the ‘perfect free kick’ by looking at the projectile motion of a football and the optimum position to take a shot at goal. By creating a linear equation for the projectile of the football’s flight path we can develop individual equations for each component that contributes to the free kick and with these equations we can see what the best technique for each step of taking a free kick is e.g. the force behind the kick.

Table of Contents

Abstract ii
Table of Contents iii
Introduction 1
Methods 4
Multiple Linear Regression 4
Data Processing 5
Results & Analysis 7
Data and Residual
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The curve created from this will create the optimum goal scoring opportunity, and any point in between will still follow the same curve pattern as the optimum one. After working through this method I decided that the position the kick was taken was so dependent on the footballers strengths and weaknesses and the defending ability of the opposition that looking at data already available to me would be more beneficial.
To do this I looked into the top 5 goal scorers of this season and in which areas of the pitch they scored from. The top 5 goal scorers this season in the premiership are : Aguero, Kane, Costa, Austin and Sanchez(according to premierleague site). I have analysed where each of their goals from this season were taken from individually and then put the data together to provide an ‘optimum area’, with the areas shown on the below diagram. Scale 1cm:10m

The data shows that the optimum place to score a goal is centrally outside the penalty box, closely followed by the Left and Right midfield but no further away than 30 yards from the
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At time t=0 ẋ = Vo CosѲ ẏ=Vo SinѲ x=0 , y=0 therefore; x=(V0CosѲ)t y=(V(little0)sinѲ) t gt solve for t then substitute into y=( Vo sinѲ) t gt to get the equation of trajectory. y= xtanѲ g

Projectile WITH DRAG m ẍ=-cẋ mӱ = -cẏ-mg c is a constant and k=c/m, m, ӱ, ẍ and g are the same as when there was no drag.
PICTURE.
Divide the equations by m and then rearrange to get; ẍ+kẋ =0 ӱ+kẏ=-g where k is the linear drag.
CALCULATING THE DRAG FORCE.
K=h(v) has a very complex variation and can only be determined experimentally so we are going to use the approximation : h(v) = (c1 +c2 |v|) V0 where v is the instantaneous projectile velocity (x1+y1), c1 +c2 are constants given by c1 = 1.55x10-4, c2 = 0.22d2 where d is the diameter of the projectile c2 >>c1 hence c1 ≈ 0 after changing h(v) to be a non-linear equation, drag is now represented as
K=
We can now use this here ẍ+kẋ =0 to make
X(t)=A+B
A and B are constants which can be obtained from the initial conditions x(0)=0 and ẋ(0)= v0 Cos therefore; ẍ(t) = the same process is used with y; y(t)= C+D -
Initial conditions y(0) = 0, and y(0) = v0 SinѲ
The expression y(t) becomes