## Bad physics in the comics…

Posting has been kind of quiet lately, because I’ve been preparing not only for my Ph.D. defence on the 20th of July, but also my move to Sydney where I’ll be taking up a post doc.  But I still have plenty to say, and I thought I would start to clear out the backlog of things I wanted to talk about.

First up on the list:  some bad physics in a comic book.  I’ve been reading old issues of The Hulk that are lying around, and I saw this set of panels that caught my eye (The Incredible Hulk #170, Dec. 1973):

Hmm. Something doesn't seem quite right...

Now, I’m not a physicist, but something doesn’t quite work here.  For one, acceleration in this situation wouldn’t be constant – it would be subject to resistance from the air, so that speed would top out at the Hulk’s terminal velocity.  From Wikipedia, the equation for terminal velocity is $V_t = \sqrt{\frac{2mg}{\rho AC_d}}$.  We’ll use some approximations for the Hulk’s weight (635 kilograms at his heaviest) and area (if we assume he’s falling perpendicular to the ground and that his width is about a third of his height, that gives an area of about 8′ * 2.24′ = 1.66 metres squared), and use a drag coefficient of 1.28 for a flat plate perpendicular to flow (since he seems to be falling parallel to the earth and not rolling into a ball or such).  The density of air at sea level is about 1.22 kg/m^3, and so using all that we can come up with an approximate terminal velocity here:

$V_t = \sqrt{\frac{2 \cdot 635 kg \cdot 9.81 m/s^2}{1.22 kg/m^3 \cdot 1.66 m^2 * 1.28}} = 69.33 m/s$

(The density of air changes with altitude, but this is fine for an informal look at the problem…)

Back in the original units of the comic, this is about 155 miles per hour, not 11,000.  (Did no-one notice that at 11,000 miles per hour, Banner would have covered the 8 miles in less than 5 seconds, even accounting for the acceleration?)  Even hitting the ground at 155 miles per hour would generate a fair amount of energy, though.  Using the work-energy theorem, we can take a rough guess at how much:

$W = \frac{1}{2} mv^2 = 0.5 \cdot 635 kg \cdot (69.33 m/s)^2$ or 1526111  joules.

Assuming (very inaccurately) that this energy would be translated into an explosion, we can guess at the TNT equivalent of the impact and get a value of 1526111 J / 4184 J/g = about 364 grams of TNT.  Not something I’d want to be sitting on top of, but not a world-breaker.

What about if he’d actually been falling at 11,000 miles per hour?  Well, that’s about 4917 metres per second. Running the same calculation as above, we get 7676162257 J, or 1834646 g or about 1.83 tons of TNT, in the range of a bunker-buster bomb – much less exciting to be sitting on top of.  The Hulk would probably be okay, but whoever he was carrying (and I forget who that is), would likely be a fine pink mist.

Of course, I’m not a physicist, and I’m working off of a vague memory of undergraduate physics courses here.  I welcome corrections from more savvy readers!  And I have to cut the writers a little slack – after all, it was 1973, and it’s not like they could just pop open their web browser and take a look at Wikipedia.