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The Bourne Identity Fall
Nov. 2nd, 2008 09:42 pmYou know the one. The fall that happens in the movie at about 1:46:41-47. I got a geeky burr under my sadle today and I calculated out the distance needed for such a fall by solving for X where (2*x/9.8m)^0.5 = 6 seconds. If I did everything right that's ((6^2)*9.8)/2 = x or 176.4 meters. Yowza. And I cannot be the only person who's done this.
Even if I'm generous and discount the first second and the last as error on my part in watching the movie that's still going to be about 78.4 meters. Quite a bit more than I can see him falling. I think a fall time of about 1.5-2 seconds would be (11-19 meters, or 4-6 floors) would be more like what real world physics would allow. Not to mention surviving a fall of that distance.
Even if I'm generous and discount the first second and the last as error on my part in watching the movie that's still going to be about 78.4 meters. Quite a bit more than I can see him falling. I think a fall time of about 1.5-2 seconds would be (11-19 meters, or 4-6 floors) would be more like what real world physics would allow. Not to mention surviving a fall of that distance.