I believe we have a Super Bowl coming up. Or, if the NFL is so picky about the use of their trademarks, I believe we have a “Big Game” coming up. As a native south Louisianian, I’m for the eternally long-suffering Saints, who in all their years have never even been to a Super Bowl. That hypothetical situation was really a part of New Orleans culture – instead of “when hell freezes over” it was always “when the Saints win the Super Bowl”. Maybe they’ll finally do it. I’m not holding my breath, but honestly New Orleans is pretty much crazy with joy that they actually made it to the game.
We can add some physics to the mix, too. Let’s figure out a football’s trajectory in terms of the initial throw. We know that as soon as the ball leaves the player’s hand there’s only one force acting on it – gravity. Well, there’s air resistance too, but a football is pretty aerodynamic so we’ll just ignore it as a first approximation. Gravity acts vertically downward, so the ball will begin accelerating straight down. But there’s no forces acting in the horizontal direction, so according to Newton there’s no change in horizontal velocity. This constant sideways velocity combined with an accelerating downward velocity gives us the arc that we’re all so familiar with.
Some of the energy of the initial throw will give the ball an initial upward velocity, and some will give it an initial horizontal velocity. By some classic trig, the first is the speed of the throw times the sine of the angle of the throw, and the second is the speed times the cosine of the angle. Write these down:
Here x0 and y0 are the initial coordinates of the ball; we’ll let them both equal 0 for convenience. Theta is the angle of the throw, v0 is the initial speed. a is the acceleration due to gravity, which is -g, or -9.8 m/s^2. The minus is because it’s accelerating downward.
Both of these equations contain the time t, which means they describe the x and y positions as a function of time after the throw. If we’re only interested in total range, we can solve one of them for t and plug it into the other. At the end of the football’s arc, it’s y position will be 0. So set y = 0, solve the y equation for t, and plug that into the x-equation. That gives you the x-position at the time at which the football has hit the ground, and we’ll go ahead and call the range R:
The usual thing to do is simplify this a little with a trig identity:
The max the sin term can possibly be is 1, for a throw at a 45 degree angle, as we might have suspected. Now you might notice that the range is proportional to the square of the initial velocity. Throw twice as fast, the ball goes four times as far. Plugging in the numbers, even at a 45 degree angle you’d have to throw at about 30 m/s = 67 miles per hour to throw the 100 yard length of the field. Short passes are much easier, but they’re thrown fast anyway to make up for the fact that a 45 degree angle throw is usually a suicidal move at short range. Lower angles and correspondingly higher speeds are required.