A daredevil plans to bungee-jump from a hot-air balloon 65.0 mabove a carnival m

Question

A daredevil plans to bungee-jump from a hot-air balloon 65.0 mabove a carnival midway. He will use a uniform elastic cord, tiedto a harness around his body to stop his fall at a point 10.0 mabove the ground. Model his body as a particle and the cord ashaving negligible mass and obeying Hooke's law. In a preliminarytest, hanging at rest from a 5.00m length cord the daredevil findshis body weight stretches the cord by 1.50m. He intends to dropfrom rest at the point where the top end of a longer section of thecord is attached to the stationary hot air balloon.
a) what length of cord should he use?
b) what maximum acceleration will he experience

Explanation / Answer

a.
For the cord, k = mg/1.5, where m = mass of the jumper.
Let L = length of the cord required.
When he falls through the height L, he loses PE = mgL and the cordbecomes taut. Thereafter, it starts stretching and stretches tillthe PE lost by the jumper is stored in the cord in the form of PEof the cord.
Let the cord stretch by x meter
=> mgL = (1/2)kx^2 and L+x = 65 - 10 = 55
=> mgL = (1/2)*(mg/1.5)*(55-L)^2
=> 3L = L^2 - 110L + 3025
=> L^2 - 113L + 3025 = 0
=> L = (1/2)[113 ± (113^2 - 4*3025)]
=> L = (1/2)[113 ± 25.865] = 69.4 or 43.6 m
=> length of cord should be 43.6 m.

b.
At the lowest point, restoring force in the cord in the upwarddirection
= k*elongation of the cord
= (mg/1.5)*(55-43.6)
= mg * (11.4/1.5)
Downward force = mg
Net force upwards = mg*(11.4/1.5) - mg = mg(9.9/1.5)
Maxm. acceleration upwards
= Net Force/mass
= mg(9.9/1.5)/m
= 6.6g m/s^2
= 6.6 * 9.8 m/s^2
= 64.68 m/s^2.

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