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Lacey asked in Science & MathematicsPhysics · 8 months ago

A daredevil wishes to bungee-jump from a hot-air balloon 70.5 m above a carnival midway.?

Update:

He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point 15.0 m above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a 5.00-m length of the cord, the jumper finds that his body weight stretches it by 1.65 m.

Update 2:

He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

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(a) What length of cord should he use?

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(b) What maximum acceleration will he experience?

3 Answers

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  • Whome
    Lv 7
    8 months ago
    Favorite Answer

    Use work energy theory.

    Assume the balloon is stationary during the event.

    from the preliminary test, we can find the spring constant k by summing forces to zero.

    kx - mg = 0

    kx = mg

    k = mg/x

    k = mg/1.65

    as this was a 5 meter cord when relaxed, we can determine a k' for a unit length of cord

    k' = 5(mg/1.65)

    The maximum distance of fall desired is

    70.5 - 15.0 - 55.5 m

    during the jump, the potential energy will convert to spring potential

    PE = PS

    mgh = ½kx²

    if the initial length of cord is L

    x = 55.5 - L

    and our spring constant will be

    k = k'/L

    55.5mg = ½(k' / L)(55.5 - L)²

    55.5mg = ½(3.03mg) / L)(55.5 - L)²

    55.5 = ½(3.03) / L)(55.5 - L)²

    36.63 = (1 / L)(55.5 - L)²

    36.63L = 3,080.25 - 111L + L²

    0 = 3,080.25 - 147.63L + L²

    quadratic formula

    L = (147.63 ±√(147.63² - 4(1)(3080.25))) / (2(1))

    L = 25.148 m

    or

    L = 122.5 m which we ignore as it cannot function in this scenario.

    L = 25.1 m ANSWER

    With the length of cord set, the stretch will be

    x = 55.5 - 25.1 = 30.4 m

    The force applied by the cord at max stretch is

    F = kx

    F = (k'/25.1)(30.4)

    F = (5(mg/1.65)/25.1)(30.4)

    F = 3.67mg

    so the maximum acceleration experienced by the jumper will be

    a = F/m

    a = 3.67 mg / m

    a = 3.67 g's

    while an observer on the ground will see the jumper's body appear to accelerate upward at one g less or

    a = 2.67 g or about 26.2 m/s²

    I hope this helps

    Please select as Best Answer if it does.

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  • audrey
    Lv 7
    8 months ago

    Ok. Does he want to win a teddy bear? What's your question?

    • ...Show all comments
    • Whome
      Lv 7
      8 months agoReport

      Yahoo limits the title length, you may wax eloquent on the question area. Maybe you should look for that area. Updates make for a very chopped up question.

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  • JetDoc
    Lv 7
    8 months ago

    Yes... AND? What's your question?

    • Lacey8 months agoReport

      I had to post it then add updates. This site only allows you to type so much in the question box.

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