I am in Florida for a meeting this week, having flown from Albany to Ft. Lauderdale. Due to the vagaries of the air travel system, though, this required a change of planes in Orlando. The Orlando-Ft. Lauderdale flight is sufficiently short that I like to think of it as a ballistic route– you’re not cruising at altitude for any significant time, but start going down practically as soon as you’re finished going up.

Let’s imagine that we have a commercial enterprise– let’s call it Angry Birds Airlines– that wants to fling objects from Orlando to Ft. Lauderdale. What would that require? That is, could you build some sort of giant catapult to fling people from Orlando to Ft. Lauderdale? And could you estimate the reasonableness of this using physics from first-year classical mechanics?

For the sake of argument, let’s imagine that the giant slingshot launcher and the glass-and-steel arrival terminal the passengers will be crashing into are at the same elevation. We can further simplify the problem by ignoring the effect of air resistance. That’s not remotely a reasonable assumption, of course, but it will do for these purposes– if the idea isn’t feasible without air resistance, there’s no way it will work with air resistance.

Given those assumptions, it’s easy to find the range for a projectile launched at speed *v* and at an angle θ with the horizontal. The result, cribbing the equation graphic from that Wikipedia page, is:

Here, *d* is the distance traveled and *g* is the acceleration due to gravity, 9.8m/s^{2}. The maximum range for a given launch velocity occurs at an angle of 45 degrees, so let’s just assume that our angry but entrepreneurial avians have set their passenger launch system at that angle.

Now, we know the start and end points for our journey, namely Orlando and Ft. Lauderdale. Google Maps says these are separated by about 200 miles (give or take), so what we want to know is what speed you would need to launch a projectile at in Orlando to have it land in Ft. Lauderdale. We can re-arrange our equation to get:

v= (gd)^{1/2}

that is, the speed we want is the square root of the distance between the cities multiplied by the acceleration of gravity.

A distance of 200 miles is 300 km at the precision we’re using, and *g* is close to 10, so we have:

v = (3,000,000 m

^{2}/s^{2})^{1/2}= 1,700 m/s

That’s 5 and a bit times the speed of sound. So, yeah, that’s not going to be a real successful money maker for our furious feathered friends. On the bright side, though, the trip would only take about four minutes…

So, if you’ve ever wondered what physicists do to pass the time during long airport layovers, now you know.