Question

A bungee jumper of mass 110 kg jumps from a cliff of height 120m. The massless relaxed bungee cord has a length of 15 m. Ignore the height of the body of the jumper. K=125N/m. (a) Find the velocity of the jumper right before the bungee starts to stretch. (b) Find the distance the cord stretches.

Please explain the steps. Thank you!

Answer 1

a)

h_i = initial height of jumper at the start of jump at the top of cliff = 120 m

L = relaxed length of the cord = 15 m

h_f = final height of jumper as the cord starts to stretch = h_i - L = 120 - 15 = 105 m

v = velocity of jumper as the cord starts to stretch

Using conservation of energy

initial potential energy = final potential energy + kinetic energy

mg h_i = mg h_f + (0.5) m v²

g h_i = g h_f + (0.5) v²

(9.8 x 120) = (9.8 x 105) + (0.5) v²

v = 17.15 m/s

b)

h_i = initial height of jumper at the start of jump at the top of cliff = 120 m

L = relaxed length of the cord = 15 m

x = stretch in the cord

h_f = final height of jumper as the cord starts to stretch = h_i - (L + x) = 120 - (15 + x) = 105 - x

k = spring constant = 125 N/m

m = 110 kg

Using conservation of energy

initial potential energy = final potential energy + kinetic energy

mg h_i = mg h_f + (0.5) k x²

(110 x 9.8 x 120)= (110 x 9.8) (105 - x) + (0.5) (125) x²

x = 26.875 m/s