How High Did Jose Jump Above the Lowest Point on His Bungee Adventure?

[abeltyukov]

Homework Statement

Jose, whose mass is 90 kg, has just completed his first bungee jump and is now bouncing up and down at the end of the cord. His oscillations have an initial amplitude of 9 m and a period of 4.0 s.2. The attempt at a solution

a) The spring constant of the bungee cord is 222.0661 N/m

b) The maximum speed at which Jose is oscillating is 14.137167 m/s

c) If the damping constant due to air resistance is 6.0 kg/s, the number of oscillations which Jose makes before his amplitude has decreased to 1.0 m is 16.479

d) From what height above the lowest point did Jose jump? (in meters)I got every other part of the question right but am having trouble figuring out part d. Do I use F=ma? Any help would be appreciated.Thanks!

 

[Dick]

The kinetic energy from Jose's fall went into two places. The first is the potential energy to stretch the bungee to a new equilibrium position (the center of his oscillations). You can get that from his weight and the spring constant. The second is any kinetic energy when he is moving. You can calculate the sum of the these at any convenient point in his oscillation - maybe easiest to do it at the top or bottom where the kinetic energy vanishes.

 

[m2003]

To solve part d, you can use the equation for the potential energy of a system in simple harmonic motion: PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the equilibrium position is at the lowest point of the bungee jump, so x = 9 m.

Since PE is equal to the initial potential energy (mgh) at the top of the bungee jump, you can set the two equations equal to each other and solve for h (the height above the lowest point):

(1/2)kx^2 = mgh

h = (1/2)(k/m)x^2

Substituting in the values given:

h = (1/2)(222.0661/90)(9)^2 = 9.9221 m

Therefore, Jose jumped from a height of 9.9221 m above the lowest point.