"Bruce Willis could reach out of the window and pluck the motionless bullet from the air."

*blush* I think I just came a little. SHROOMIFIED! Send your physics teacher a thank you note for me.

He's all mine, ladies.

This is probably the best question and answer that has ever happened on Questionland. Of course, I say that being a huge Die Hard nerd.

"What's better than Die Hard? TWO DIE HARDS!!" - Joey

Hehe - I'll try not to let it go to my head, Megan.

Though if you see someone on B'way in a modified brownshirt that has added at top the words: "ASKED THE BEST QUESTION EVER ON QLAND" ... er... there's the slight chance it's me.

Or I might just change my by-line... *evil grin*

*just noticed sidebar ad that says Jon and I won free pizza*

!!wow, thanks Pagliacci!!

It looks like I got some criticism on Slog for this.

At risk of ruining the joke, "spherical villain" is just a reference to a certain spherical farm animal, not a serious premise for the analysis.

...not that any part of this whole exchange should be considered serious.

The terminal velocity given isn't actually for a spherical mass. It's 55 m/s, which is close to just about any mention on the web of terminal velocity of a randomly-oriented human.

No, it isn't an exact model of the acceleration / velocity / distance. Most importantly, acceleration would decrease with velocity due to increased drag, and at the same time would increase slightly as the distance between the attracting bodies decreases. Unlike terminal velocity, which can greatly affect a distance calculation over time, I didn't consider these two factors to be particularly more important than, say, factors due to relativity. See: spherical villain.

As for the equal and opposite force of the gunshot itself, I'm ignoring any consideration of torque on the villain and assuming one shot would be fired. If the villain did shoot with a completely rigid body, the equal and opposite work applied should be close to the cited 477 J. It's not immediately obvious to me how that would affect the net velocity of the bullet. The work would be applied to the villain over the muzzle length, and you might need calculus to derive the change in velocity. Just at a glance, adding 447 J of kinetic energy directly to an 80 kg male would only result in a velocity change of about 11 ft/s. Not enough to substantially change the analysis, and certainly not more of a change than if the villain had used hollow point ammunition, for instance.

for those not willing to sully themselves in SLOGland

@11 (reverse polarity) said:
Interesting math, but incorrect.

(1) Humans are NOT spherical. They fall at different speeds, depending on their angle to the wind (we've all seen videos of skydivers all spread out like spiders to slow their fall). The terminal velocity of a falling human is around 65 miles per hour (ask any skydiver).

(2) So even if our buddy Hans reaches terminal velocity, there is enough extra muzzle velocity in a .45 to do damage.

(3) Every action has an equal and opposite reaction. That means Hans only gets one shot while in free fall. The recoil of his first shot would set him spinning, making any subsequents shots completely randomly aimed.

And then is set straight by this post by Biffster:
Interesting corrections, but incorrect.

1) "Assume spherical villain..." If you want to do the fluid dynamics calculations to calculate the drag coefficient that Hans possesses as he falls to his doom, be our guest. If not, let's just keep assuming he's spherical.
And I asked my skydiving friend "Wikipedia," about reaching terminal velocity and this is what he had to say, "...the terminal velocity of a skydiver in a free-fall position with a semi-closed parachute is about 195 km/h (120 mph or 55 m/s) -Huang, Jian (1999). "Speed of a Skydiver (Terminal Velocity)". The Physics Factbook. Glenn Elert, Midwood High School, Brooklyn College. http://hypertextbook.com/facts/JianHuang….

2) You're making the assumption that gravity doesn't exist to slow down the ascent of the fired bullet. You're also assuming that he fires the bullet right when he reaches terminal velocity, he's not. The bullet that Bruce Willis plucks from mid-flight has been fired after just over a minute from entering free-fall, not at the moment when Hans reaches terminal velocity.

3) This would be true if Hans' arm was a rigid structure attached to his center of gravity, but it's not. Unless he was firing from his shoulder (like with a rifle) this would likely not happen. His arm would absorb most of the recoil from the pistol.

Reading comprehension is a bitch...

Thanks Rev, I didn't actually see that reply on Slog.

I don't really know what point was being made at 2. I also concluded that the bullet would still injure. Using a lower terminal velocity would just result in even more kinetic energy.

Megan-the-DieHard-nerd: if you haven't heard it/don't already own the CD, you may be interested in this little ditty:

http://www.youtube.com/watch?v=OTyw6cq86kY

This has been bothering me for a while. Unless I'm missing something in rereading my math, the villain would only have to fall for about 35 seconds for the bullet trick to work, not just over a minute (I layed it out right there in the second to last paragraph: 33.54 seconds + 5.61 seconds). Maybe a minute just sounded better at the time.