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back to boardIt seems like that the author don't know the difference between speed and engery. Edited by author 08.09.2004 19:43 Re: It seems like that the author don't know the difference between speed and engery. Posted by Vovka 21 Oct 2004 16:30 I don't think so. T ~ V^2, assuming that gives ac Re: It seems like that the author don't know the difference between speed and engery. Posted by div 13 Sep 2006 17:22 hmm... i think the same. In each jump energy decreases K times, then (E = mv^2/2) speed decreases sqrt(k) times. But my solution, which assuming that speed descreases k times (not sqrt(k)), got AC... Strange Sorry for my poor English=) Re: It seems like that the author don't know the difference between speed and engery. No,you are wrong,it's correct. Each time v decreases in sqrt(k) times,but we use Vx*Vy,so it decreases in k times. Re: It seems like that the author don't know the difference between speed and engery. I agree with you kinetic engergy decreases in K times speed decreases in sqrt(K) times Because E=1/2*m*v^2 Edited by author 05.12.2008 16:28 |
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