The Physics of the Batarang
Entry posted by mathgeek15
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Originally this was going to be about Batman's utlilty belt, but then I had so much fun things to talk about with the batarang I changed my mind.
Well, to begin with, the batarangs are NOT a sold piece of hardened steel alloy. It actually folds in half. Another thing to keep in mind, the batarangs are not designed to return to Batman, hence why the design is not very similar to a boomerang. Also, Batman has multiple kinds of batarangs. So when Batman throws one at a rope, he has one for cutting. To knock an enemy down, he has a hard-impact one. So many of you people are probably thinking by now, how does such a little piece of metal knock a grown man down?
So lets say a thug is trying to run away from batman after robbing a jewelry store, leaving the jewels behind because he's soo scared of Batman. The average mass of a 30 year-old male is about 180 lbs, or 82 kg, and the average speed of a male of this weight running is about 8 mph, or 12.87 kmph. Since I do not know the actual weight of a batarang, I am going to say it's about .5 lbs (hard impact batarangs are about twice the size of the stereotypical batarang), which would be about .227 kg. Finally, the velocity of the batarang would be about 90 mph, considering that is the typical speed of a baseball pitcher and Batman has a pretty good arm himself, making it about 144.8 kmph. Time for some Momentum!!!! This scenario is would be an inelastic collision since the man falls on his face. Thus, the resulting velocity needs to be high enough to make the man lose his balance.
MTVT+ MBVB= (MT+MB) V
(82)(12.87)+(.227)(144.8)= (82+ .227) V
+ 32.87= 82.227V
V=13.23 kmph
Ok, even when I looked at that I thought something was wrong. But even this change in velocity will throw a person off balance because of the immediate change. Then Batman gets to walk over and handcuff him or tie him up and hang him like a pinata. Anyone got a stick?
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