Question

In: Physics

A bungee jumper of mass 110 kg jumps from a cliff of height 120m. The massless...

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!

Solutions

Expert Solution

a)

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

L = relaxed length of the cord = 15 m

hf = final height of jumper as the cord starts to stretch = hi - 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 hi = mg hf + (0.5) m v2

g hi = g hf + (0.5) v2

(9.8 x 120) = (9.8 x 105) + (0.5) v2

v = 17.15 m/s

b)

hi = 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

hf = final height of jumper as the cord starts to stretch = hi - (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 hi = mg hf + (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


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