Thursday, December 10, 2015

A Look at Goal Kicks in Soccer

I've been playing soccer for my whole life, so I decided to study goal kicks, specifically those taken by professional soccer players. I estimated that on average, when a goalkeeper takes a goal kick, they kick the ball 55 meters across the field at an angle of approximately 45o. With this information, I decided to use physics to determine the initial velocity that the ball must be kicked at in order to reach 55 meters. I was also able to find the maximum hight that the ball would reach during its trajectory, as well as how long it would take for the goal kick to be completed. For the purpose of solving this problem, I ignored air resistance, and therefore this became a projectile motion problem. I also took into consideration that the average length of a professional soccer field is 110 metersI began this problem by writing down the X and Y components:

 X components Y components a = 0 m/s2 a = -9.8 m/s2 x = 55 m y = ? vx = vcos45o vy = vsin45o t = ? t = ?

Since this is a projectile motion problem with no air resistance, we know that the ball will hit the ground with the same velocity as its initial velocity, but in the downwards direction. Therefore, the initial and final velocities in both the x and y directions will have the same magnitude. Knowing this, I used the Y components to solve for the velocity of the ball in terms of time:

vf = vo + at
-vsin45o = vsin45o + (-9.8 m/s2)t
-2vsin45o = -9.8t
t = (-2v(.7))/(-9.8)
t = .14v s

Then I used this answer, in addition to the X components, to find the velocity of the ball:

x = vot + ½at2
55 m = (vcos45o m/s)(.14v s) + ½(0 m/s2)(.14v s)
55 = v2(.098)
v= 561.22
v = 23.7 m/s

Therefore, the time of the soccer ball’s trajectory would be:

t =.14v
t =.14(23.7 m/s)
t = 3.32 s

Then I separated the velocity into its X and Y components:

X: 23.7cos45o = 16.59 m/s
Y: 23.7sin45o = 16.59 m/s

Using this information, I found the maximum height that the soccer ball would reach during its trajectory:

vf2 = vo2 + 2ay
(0 m/s)2 = (16.59 m/s)2 + 2(-9.8 m/s2)y
0 = 275.23 – 19.6y
y = (-275.23 / -19.6)
y = 14 m