The idea came for this post came from a social media post regarding a game winning field goal from the Eagles vs. Giants game this past Sunday. With no time left on the game clock, Jake Elliot kicked a field goal from 61 yards (tied for 7th best all-time), and just narrowly managed to squeeze the ball inside the right upright and over the bottom goal post. The post showed that Elliot had to kick the ball within a tight 5.8 degree angle. It related this to being the equivalent of kicking the ball into an 18 inch window from 5 yards out. For this blog post, I decided to look at both the accuracy of that post and how much speed he had to generate to put enough air under the ball for it to go the distance. Elliot kicked the ball from the the right hash, conveniently at a right angle to the plane formed by the goal posts (the hash marks and uprights are both 18 feet, 6 inches apart). This means that the tangent of the angle between which Elliot could have kicked the ball to win the game was equal to 18.5 ft divided by 183 ft (61 yds times 3 ft per yard). Upon checking, report accurately determined the angle, (5.77 degrees to be just a little more precise.) A little more trig shows that x/15ft=tan(5.8). 15*tan(5.8)=1.52, meaning that the report again was accurate in that from 15 feet away, the ball would have to squeeze into about an 18 inch window to be on target when it reached the goal posts! Pretty remarkable when you think about the fact that an NFL ball is 22 inches in circumference at its fattest point. Further, in order to determine the initial velocity he had to put on ball to get it over the 10 foot high goal post, via a direct line drive at the shortest possible distance (just over 183 ft). In this case again, we can look at right triangles to see that our total distance on the ground is 183 ft, and our elevation is at 10 ft. This means again that the inverse tangent of the height over the ground distance is equal to the hypotenuse, or the total distance the ball must travel. This comes out to a 3.1 degree angle if Elliot were to kick the ball so that it just skimmed the bottom right corner of the goal post on a straight line. Now using vf^2=vo^2+2a(xf-xi) we can determine what the exit velocity in the y direction would have to be, which we can use to find the overall initial velocity. vf= 0 at the peak of a curve, and vi is our unknown. The only acceleration on the ball after it is kicked is gravity at -9.8 m/s, and our xf=3.048 m (10 ft). This means that the ball must exit at 7.7 m/s in the y direction only. If you plug that into a sin(3.1 degrees)=7.7/hypotenuse, you can see that the initial velocity would be at least 142 m/s!!! Obviously, I have not taken wind resistance/drag into account, which does play a role no doubt, but the feat is impressive nonetheless. In conclusion, Jake Elliot has a hammer for a foot,

and lets go Birds!!!!!!

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