At the 1939 World’s Fair, San Francisco Seals catcher Joe Sprinz tried to catch a baseball dropped from the Goodyear blimp 1,200 feet overhead.

Sprinz knew baseball but he hadn’t studied physics — he lost five teeth and spent three months in the hospital with a fractured jaw.

- from Futility Closet

Futility Closet is one of my new obsessions: hundreds upon hundreds of logic puzzles, oddities, surprises and more (the podcast is one of the highlights of my week). But they’ve done something I frown on, here, and begged a very important question: how fast was the ball going when Sprinz caught it?

In basic mechanics, we ignore air resistance, even though it’s going to be significant in this case. We’ve got a ball dropped from 1200 feet (around 370 metres, rounding sensibly) under gravity.

You’ve got your suvat equations: we know $u=0$, $s=370$, $a = -9.8$ and want to know $v$, so we’ll simply say $v^2 = u^2 + 2as$, or $v = -\sqrt{2\times 9.8 \times 370} \simeq - \sqrt{7250}$ or somewhere about 85 metres per second straight down. What’s that in sensible speeds? In kilometres per hour… it’s a little more than 300.

In terms of kinetic energy, a baseball has a mass of about 0.15kg, so we’d need to do something around a kilojoule of work to stop it - about the same as an 80kg runner going at 3.6 m/s (or 13 km/h, enough for a very respectable marathon time. If I ran into a wall, even at my slightly more ponderous pace, I’d definitely know about it - and losing a few teeth is certainly plausible!

Now, I’ve not taken into account the effects of air resistance, which will certainly slow the ball down: that’s because the equation gets complicated! The resistance force is $\frac 12 \rho v^2 C_D A$, where $\rho$ is the density of the air (which varies with height and temperature), $v$ is the velocity, $C_D$ is the drag coefficient (about 0.3 for a baseball) and $A$ is the area (about 0.02 m²).

At a normal temperature at sea level, $\rho$ is about 1.3 kg/m³, so we’d need to work out something like:

$F = (0.15)a = -g + \frac 12 (1.3 )(0.3)(0.02) v^2 \simeq -g + 0.004 v^2$

… and there’s no obvious way (to me) to integrate that. (Oddly, it’s the constant $g$ that’s the problem; otherwise, it’s a pretty standard FP2 integrating factor problem. If you know how to do it, let it be known in the comments!)

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## MathbloggingAll

A baseball with your name on it http://t.co/rW58GoB6fL

## srcav

Great post from @icecolbeveridge unfortunately I can’t help in this case! http://t.co/ALebI0oKVn