Drop Kick Vs. Punt
Which is Better?

By: Jack Seifert

Within goalkeeping, there is a ongoing debate of whether to use a drop kick or a punt when kicking the ball far up field. I have had coaches tell me arguments for both, and that is why for my project I decided to find out what one is the best based on physics.  

A drop kick is when you let the ball hit the ground first and strike the ball as it bounces up off of the ground. A punt is when you drop the ball in front and to the side of you and strike it on the way down

What am I measuring and how am I going to do this?

To figure out what one is better I will be analyzing and measuring a few things. First I will take a video of my drop kick and my punt, put the videos into logger pro and analyze the flights of the kicks. What one goes farther?  I also will measure the average drag on each ball in the X direction. I want to see how much force actually acts upon the ball, and if there was no drag how much farther would it actually go.

Path of a drop kick versus a punt

A path of a drop kick is much flatter then a punt. The ball travels at a lower trajectory, and then at the end of the kick when the starts traveling back towards the ground, it falls in a steep arc. While with a punt, the ball stays in a relatively constant arc that goes higher then the drop kick.

Energy of each kick

It is quite obvious that each type of kick give the ball quite a lot of energy at the beginning of its flight, but how much energy does the ball lose at the end of its flight for each type of kick?

Drop Kick

Initial kinetic energy of ball = .5 x mass of ball x (initial X-velocity)^2

= .5 x .432Kg x (27.431m/s)^2

= 162.531 J

Final kinetic energy of ball = .5 x mass of ball x (final X-velocity)^2

= .5 x .432Kg x (9.271m/s)^2

= 18.5615 J

During the flight of the ball, the ball lost 143.97 J of energy.

Punt

Initial kinetic energy of ball = .5 x mass of ball x (initial X-velocity)^2

= .5 x .432Kg x (19.66m/s)^2

= 83.4959 J

Final kinetic energy of ball = .5 x mass of ball x (final X-velocity)^2

= .5 x .432Kg x (13.4m/s)^2

= 38.7944 J

During the flight of the ball, the ball lost 44.7015 J of energy.

What is causing these massive losses in energy you might ask. Well, the answer is the force

Drag, What is it?

Drag is a force that opposes motion. The direction of the drag force is always opposite the direction of the objects motion. The equation for drag is:

Where p is the density of air, v is the velocity, Cd is the drag coefficient, and A is the cross sectional area of the object. Since drag is a force it is measured in newtons.

For my project I will only be solving for the drag acting on the ball in the X direction.

Drag on Drop kick

Using the energy lost during a drop kick I solved for above, I can solve for the average force  of drag. I can also solve for the average force of drag using the drag model from above.

143.97 Nm = Fdrag x 50.9m

Avg. Fdrag = 2.83 N when solving using energy

Avg. Fdrag = .5 x p x v average^2 x Cd x A

         = .5 x 1.2 kg/m^3 x (14.67m/s)^2 x .47 x .038m^2

Avg. Fdrag = 2.30574 N when solving using the drag model            

Drag on a punt

Using the energy lost during a drop kick I solved for above, I can solve for the average force of drag. I can also solve for the average force of drag using the drag model from above.

44.7015 Nm = Fdrag x 45.56m

Avg. Fdrag = .981157 N when solving using energy



Avg. Fdrag = .5 x p x average v^2 x Cd x A

= .5 x 1.2 kg/m^3 x (15.1867m/s)^2 x .47 x .038m^2

Avg. Fdrag = 2.47149 N when solving with the drag model

Summary

As you can tell, the forces of drag using the two different methods of solving for it are not that close to each other. That is because of one big reason. The model of drag I used is not very accurate. To have an accurate model many other factors have to be taken into consideration such as what material the soccer ball is made out of, the aerodynamics of the soccer ball due to the type of paneling used on the cover, the spin on the ball and a soccer ball is far from a perfect sphere. The drag causes a much larger decrease in energy to the drop kick in the x direction. That is because within a drop kick most of the energy is in the x-direction, and very little in the y- direction. So more energy will be lost in the x-direction when compared to the punt who has less energy in the x-direction, but more in the y-direction, because as I said before, the arc of a punt is much higher then a drop kick.

So what one will it be, Drop kick or Punt?

In a perfect world with no drag, I would go with a drop kick. That is because according to the projectile motion model, the x-velocity remains constant, which means in the 3.47 second test period I had, my drop kick would go 95.2 meters. When in the same period my punt would only go 68.22 meters. But because we have to live in a non perfect world I would have to choose a punt. Because in the 3.47 second test period for my drop kick, it only went 50.9 meters because of drag. My punt, in the same amount of time with drag would go 52.7 meters. It may seem like a small distance, but in the game of soccer, those two meters could be the difference between getting it to your teammate for a chance to score, and hitting it two meters short of your teammate to the other team, and them coming right back at you on offense.

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