Hey guys! With Superbowl XLV quickly approaching (and MCATS…) I thought it would be neat to talk a little about a single aspect of the game that’s rarely discussed among sports fans – the science. I’ll be starting a series of threads each detailing some of the quintessential roles that science plays in the sport, primarily in the fields of physics, biochemistry, and biology. These threads will be informational at the very least and will serve as a useful tool for me to review for the MCAT. Also, I thought I’d share some info with you guys because its actually pretty interesting stuff. So when you’re watching the big game, put down the Doritos and take a second to think about all of the crazy shit that’s happening right in front of you… you’re mind might just fry.

So on today’s menu: Physicsicon_twisted.gif

During a football game (or any other type of game for that matter) there is actually quite a lot of physics going on; definitely more than I can talk about here. But I’ll try to hit some of the highlights and keep discussion restricted to things I think people would be interested in.


Being a punter must be a nerve-wrecking job. The game halts and millions of people all over the world watch you as you make that single field goal kick. One slip and millions of people watching instantly transform into millions of angry fans. But if he knows his physics he’ll be sure to no disappoint. When the punter kicks the football, there are three variables that he has control of:

• The velocity at which the ball leaves his foot
• The angle of the kick (we’ll call this theta – θ)
• The rotation of the football

The rotation of the football influences how the ball will slow down in flight because the ball is affected by drag caused by air resistance. A spiriling ball will have less drag so it doesn’t slow down as much and thus goes further. It is possible to calculate the influence that air resistance has on a ball but its actually quite complex and a little beyond the scope of this thread. Unless you’re doing rocket science most physicists will deem air resistance as negligible.

The velocity of the ball and the angle (θ) are major factors that determine how long the ball will remain in the air (hang time), how high the ball will go, and how far the ball will go.

When the ball is kicked, it leaves the foot traveling in a parabolic trajectory. Being a parabola, it is easily separated into its two principle components: X (vertical) and Y (horizontal). How fast the ball will travel in either direction depends not on the force with which the ball is kicked but rather the angle in which it’s kicked (weird eh?). If the angle is steep, a ball will have more velocity in the vertical direction than the horizontal. It will travel higher, have a longer hang time, but it will only travel a short distance in the horizontal direction. If the angle is shallow, the ball will have more velocity in the horizontal direction but it will not travel very high, it will have a short hang time, and it will travel a long distance in the horizontal axis. These are important factors to be considered by the punter.

So, now lets delve into the nitty gritty…

The parabolic path of a football is easily described with two equations:

• Y= Vy(t) – 0.5gt^2
• X = Vx(t)

Now at first glance equations can be intimidationg, but they aren’t so bad if you know what all the variables represent.

Y = height of ball at any time (t)
Vy = the vertical component of the footballs initial velocity
g = acceleration due to gravity – 9.8 m/s^2

-As a side note, its important to know that when an object is falling on a planet with gravity, it never falls at a constant speed. In fact, it accelerates until it something stops it or it reaches terminal velocity – the velocity at which no more acceleration can occur because of air resistance. On earth, objects accelerate downward as they fall at a rate of 9.8 m/s^2.

X = the horizontal distance of the ball at any time (t)
Vx = the horizontal component of the footballs initial velocity

See now its not so bad is it?

If we were to know the exact velocity (V) of the kickers foot and the angle (θ) at which he made contact with the ball we could very accurately calculate the balls hang time (Ttotal), the balls peak height (Ymax), and the balls maximum range (Xmax). Cool right? Now I know you’re all dying to know how to do it… you’re going to need to do the following:

Break the velocity down into its X & Y components using the following formulas:

• Vx = V cos (θ)
• Yy = V sin (θ)

…ZOMG COS AND SIN… WHAT THE HELL ARE THOSE!? They are triginometry functions… don’t worry they are standard on every scientific calculator, you don’t actually have to know how to calculate them.

Hang time (Ttotal) is determined by one of the following two formulas. It doesn’t really matter which one you use its just preference.

• Ttotal = (2Vy / g)
• Ttotal = (0.204)(Vy)

The 0.204 in the second equation is a constant value, its not something that we derive.

Once you know the hang time and the Vx component, you can calculate max range.

• Xmax = (Vx)(Ttotal)

Next you can calculate at which time during its trajectory does the ball reach its peak height.

• T1/2 = (0.5)(Ttotal)

And finally with all the numbers obtained you can calculate the peak height using once of two formulas – again based on preference.

• Ymax = [(Vy)(t1/2)] – 1/2g(t1/2)^2
• Ymax = [(Vy)(t1/2)] - 0.49(t1/2)^2

Ok, now that we know what we’re doing lets try a real example. Say a punter makes a field goal kick and the ball left his foot traveling at 26.3m/s and θ was 29°.

First break down into X & Y components:

Vx = (V) cos 29 = (26.3m/s)(.874) = 22.9 m/s
Vy = (V) sin 29 = (26.3 m/s)(.484) = 12.7 m/s

Determine hang time

Ttotal = (0.204)(12.7) = 2.59 m/s

Determine Max Range

Xmax = (22.9m/s)(25.9 m/s) = 59.31 M

Oh noes… silly physicists, THIS IS AMERICA! We don’t play our games with silly meters. We like to make things complicated and use yards ;)

So… since 1 meter = 1.09 yards then Xmax = 64.64 yards.

Determine at how many seconds is peak height reached

T1/2 = (0.5)(2.59 m/s) = 1.295 s

And finally you can find the peak height

Ymax = [(12.7 m/s)(1.295)] – 0.49 (1.295)^2 = (16.44) – (.8217) = 15.61 M

No meters in American football so since 1 meter = 3.28 feet… Ymax = 51.22ft

So that’s the physics of punting in a nutshell. Kudos if you read through the whole thing. Hope it was an enjoyable (or at least informative) read. Questions and comments welcome!