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Calculating Bungee Jumping Acceleration: Physics and Calculus Problem Solutions
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[QUOTE="Simon Bridge, post: 5476353, member: 367532"] I'll annotate the problem statement ... this is how I'm reading it. The bungee is stretched at the equilibrium point... otherwise how is it an "equilibrium"? You have F=ks where s is the extension and k=0.39 slugs/s[sup]2[/sup] (guessing units) The equilibrium point is where this force matches the weight of the jumper. Your reasoning is that the jumper loses gravitational potential energy, exchanging for kinetic energy (and work to overcome air resistance) until the unstretched length is reached (is terminal velocity reached first?) ... then kinetic and potential energy is exchanged for extension in the chord until there is none left. ... oh I hate doing things in negative numbers: I always lose a minus sign somewhere. So v is specified for a particular position? That would be useful. ... so if the position of the jumper is h, then h(0)=-L and ##\dot h(0)=-113##ft/s ... are they expecting you to solve a second order differential equation. ... and changing variables again! Oh well... they tell you that ##H(0)=-L##, ##V(0)=\dot H(0)## which you chose when you picked the chord, and ##A(0) = ## from problem 1. [B][/B] You are given ##v=\dot h## here. For t<0 ##\dot H - mg = m\ddot H## ... second order DE. Fortunately you don't have to solve it (yet) ... ##A(0)=\frac{1}{m}(V(0)-mg)## fwiw ... that's how I'm reading it. [/QUOTE]
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Calculating Bungee Jumping Acceleration: Physics and Calculus Problem Solutions
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