A person is going bungy jumping. there is 2 kinds. Wet and Dry. in wet the person is submerged to a depth of about 1m. in dry, the person fall ends jsut above the surface of water. In one case, the bridge os 43m above the river and in the other case its 71m above the river. Participants jump fromt he bridge, fastened to an elastic rope which is adjuseted to halt their descent at the appropriate level. The rope is specially designed and its modulis of elasticity is known from specifications. For the purpose of this problem, we will assume th rope is streched to twice its natural length bu a person of mass 75kg hanging at the rest from the free end. It is necessary to adjust the length of the rope in terms of th weight of the jumper. 1) For a preson of mass m kg, calculate the depth of to which the person would fall if attached to a rope of the type described above, with length (l) metres. 2) FOr the 71m jump, aompare (l) as a fucntion of m for the following twwo jumps: A dry jump, where the descent is halted 1m above the water and a wet jump, where the descent is halted below the surface. equations that maybe of use: T = kx (T=tension, k= is know as the stiffness or spring constant, x= extension beyond natural length) T = (lander / l)x (lander isthe modulus of elasticity and l = natural length) Force = mass x accerleration Also derivatives and integration is used.