Design a “bungee jump” apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 15 m long, and that the cords stretch in the jump an additional 20 m for a jumper whose mass is 100 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground).(e) Which of the following statements is a valid basis for answering part (d) correctly? [Part(d) answer was the jumper’s momentum is changing]Since the net force must be zero when the momentum is zero, and since dpy/dt is equal to the net force, dpy/dt must be zero.A very short time ago the momentum was downward (and nonzero).If the momentum weren’t changing, the momentum would remain zero forever.After a very short time the momentum will be upward (and nonzero).Since the momentum is zero, the momentum isn’t changing.Check to make sure that the magnitudes of the velocity and force vectors shown in your diagram number 4 are consistent with your analysis of parts (c), (d), and (e).(f) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness ks for each of the two cords.N/m (g) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.)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.)g’s (acceleration in m/s2 divided by 9.8 m/s2)