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# Physics Question?

See picture for first part of problem and pictures.

II. You are to find:

a) The necessary un-stretched length of the bungee cord.

b) The maximum speed and acceleration achieved during the jump.

c) The spring constant of the bungee cord in relation to the jumper’s

mass.

### 1 Answer

- NCSLv 75 months agoFavorite Answer
Welcome to Yahoo!Answers.

All quantities are in SI units.

person's GPE becomes spring PE:

m*g*(25 - 2) = ½kx²

where x is the maximum elongation of the cord,

h - 2 = L + x

where L is the unstretched length

At equilibrium, the stretch z is due only to the person's weight:

mg = kz

and

x = z + 8

so

mg = k(x - 8)

k = mg / (x-8)

m*g*23 = ½*[m*g/(x-8)]*x²

m*g cancels

23*2 = x² / (x-8)

46(x - 8) = x²

x² - 46x + 368 = 0

is quadratic with roots at

x = 10.3 m ← solution

and x = 35.7 m ← outside of acceptable range

a) so

L = 23m - x = 12.7 m

b) the maximum speed is achieved just when the spring force is equal to the weight:

z = x - 8m = 2.3 m

KE = initial GPE - final GPE - SPE

KE = m*g*(12.7+2.3) - ½*[m*g/2.3]*(2.3)²

KE = m*g*(15 - 1.15) = m*g*14

and of course KE = ½mv²

so after canceling the mass we have

½v² = g*14 = 9.8*14 = 137

v² = 274

v = 17 m/s

c) k = mg / (x - 8) = m*9.8 / 2.3 = 4.26*m

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