Design a \"bungee jump\" apparatus for adults. A bungee jumperfalls from a high platform with two elastic cords tied to theankles. The jumper falls freely for a while, with the cords slack.Then the jumper falls an additional distance with the cordsincreasingly tense. Assume that you have cords that are 14 m long,and that the cords stretch in the jump an additional 20 m for ajumper whose mass is 80 kg, the heaviest adult you will allow touse your bungee jump (heavier customers would hit the ground). (a)It will help you a great deal in your analysis to make a series of5 simple diagrams, like a comic strip, showing the platform, thejumper, and the two cords at the following times in the fall andthe rebound: 1 while cords are slack (shown here as an example toget you started) 2 when the two cords are just starting to stretch3 when the two cords are half stretched 4 when the two cords arefully stretched 5 when the two cords are again half stretched, onthe way up On each diagram, draw and label vectors representing theforces acting on the jumper, and the jumper's velocity. Make therelative lengths of the vectors reflect their relative magnitudes.(b) At what instant is there the greatest tension in the cords?(How do you know?) When the person has fallen between 0 m and 14 m.When the person has fallen between 14 m and the bottom. At the top,when the person has fallen 0 m. At the bottom, when the person hasfallen 34 m. When the person has fallen 14 m. (c) What is thejumper's speed at this instant, when the tension is greatest in thecords? v= m/s (d) Is the jumper's momentum changing at this instantor not? (That is, is dpy/dt nonzero or zero?) (e) Which of thefollowing statements is a valid basis for answering part (d)correctly? A very short time ago the momentum was downward (andnonzero). Since the momentum is zero, the momentum isn't changing.If the momentum weren't changing, the momentum would remain zeroforever. Since the net force must be zero when the momentum iszero, and since dpy/dt is equal to the net force, dpy/dt must bezero. After a very short time the momentum will be upward (andnonzero). the tolerance is +/-5% (f) Focus on this instant ofgreatest tension and, starting from a fundamental principle,determine the spring stiffness ks for each of the two cords. ks=N/m (g) What is the maximum tension that each one of the two cordsmust support without breaking? (This tells you what kind of cordsyou need to buy.) FT= N (h) What is the maximum acceleration|ay|=|dvy/dt| (in \"g's\") that the jumper experiences? (Note that|dpy/dt|=m|dvy/dt| | if v is small compared to c .) |ay|= g's(acceleration in m/s2 divided by 9.8 m/s2) (i) What is thedirection of this maximum acceleration? (j) What approximations orsimplifying assumptions did you have to make in your analysis whichmight not be adequately valid? (Don't check any approximations orsimplifying assumptions which in fact have negligible effects onyour numerical results.) Assume that the gravitational force hardlychanges from the top of the jump to the bottom. Neglect airresistance, despite fairly high speeds. Assume the speeds are verysmall compared to the speed of light. Assume tension in cordproportional to stretch, even for the very large stretch occurringhere.
