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 45^{o}. 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 meters. I began this problem by writing down the X and Y components:
X components

Y components

a = 0 m/s^{2}

a = 9.8 m/s^{2}

x = 55 m

y = ?

v_{x }= vcos45^{o}

v_{y }= vsin45^{o}

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:
v_{f }= v_{o} + at
vsin45^{o} = vsin45^{o} + (9.8
m/s^{2})t
2vsin45^{o} = 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 = v_{o}t + ½at^{2}
55 m = (vcos45^{o }m/s)(.14v s) + ½(0 m/s^{2})(.14v
s)
55 = v^{2}(.098)
v^{2 }= 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: 23.7cos45^{o} = 16.59 m/s
Y: 23.7sin45^{o} =
16.59 m/s
Using this information, I found the maximum
height that the soccer ball would reach during its trajectory:
v_{f}^{2 }= v_{o}^{2}
+ 2ay
(0 m/s)^{2} = (16.59 m/s)^{2} +
2(9.8 m/s^{2})y
0 = 275.23 – 19.6y
y = (275.23 / 19.6)
y
= 14 m
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