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(a) A bungee jumper of height 2 m falls from a bridge which is 125 m above the surface of the water, as shown in the diagram - HSC - SSCE Mathematics Extension 2 - Question 7 - 2009 - Paper 1

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(a)-A-bungee-jumper-of-height-2-m-falls-from-a-bridge-which-is-125-m-above-the-surface-of-the-water,-as-shown-in-the-diagram-HSC-SSCE Mathematics Extension 2-Question 7-2009-Paper 1.png

(a) A bungee jumper of height 2 m falls from a bridge which is 125 m above the surface of the water, as shown in the diagram. The jumper's feet are tied to an elasti... show full transcript

Worked Solution & Example Answer:(a) A bungee jumper of height 2 m falls from a bridge which is 125 m above the surface of the water, as shown in the diagram - HSC - SSCE Mathematics Extension 2 - Question 7 - 2009 - Paper 1

Step 1

Show that $x = \frac{g}{r^2} \ln{\frac{g}{g - rv}} - \frac{v}{r}$

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Answer

To derive the equation from the given second-order differential equation, start by substituting the expression for acceleration.

  1. Given d2xdt2=−g−rv\frac{d^2x}{dt^2} = -g - rv, rewrite it as:

  2. Next, using the relationship between acceleration, velocity, and displacement, observe that: dvdt=dvdx⋅dxdt=vdvdx.\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}.

  3. Replacing acceleration we get: vdvdx=−g−rv.v \frac{dv}{dx} = -g - rv.

  4. Rearranging and integrating: ∫v dv=∫(−g−rv) dx. \int{v \, dv} = \int{(-g - rv) \, dx}.

  5. Solving these integrals accordingly leads to the desired expression.

  6. Finally, substitute the initial conditions of x=0x = 0 and v=0v = 0 to verify the results.

Step 2

Find the length, L, of the cord.

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Answer

Given that g=9.8 m s−2g = 9.8 \, m \, s^{-2} and r=0.2 s−1r = 0.2 \, s^{-1}, to find L such that the jumper's velocity is 30 m/s:

  1. Start with the derived expression for velocity at x=Lx = L: v=gr(1−e−rL)v = \frac{g}{r} (1 - e^{-rL}). Setting v=30v = 30 m/s gives:

    30=9.80.2(1−e−0.2L).30 = \frac{9.8}{0.2} (1 - e^{-0.2L}).

  2. Solve for LL:

    30=49(1−e−0.2L) 30 = 49 (1 - e^{-0.2L}) 1−e−0.2L=3049, 1 - e^{-0.2L} = \frac{30}{49}, thus, e−0.2L=1−3049=1949. e^{-0.2L} = 1 - \frac{30}{49} = \frac{19}{49}.

  3. By taking the natural logarithm, we find: −0.2L=ln⁡(1949) -0.2L = \ln{\left(\frac{19}{49}\right)} and solving gives: L=−ln⁡(1949)0.2.L = -\frac{\ln{\left(\frac{19}{49}\right)}}{0.2}.

  4. Calculate the numerical value and round to two significant figures.

Step 3

Determine whether or not the jumper's head stays out of the water.

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Answer

In the second stage where x>Lx > L, use the given equation:

x=et10(29sin⁡t−10cos⁡t)+92.x = e^{\frac{t}{10}}(29\sin{t} - 10\cos{t}) + 92.

  1. Determine the maximum value of the function 29sin⁡t−10cos⁡t29\sin{t} - 10\cos{t}, which is at least: 292+(−10)2=841=29.\sqrt{29^2 + (-10)^2} = \sqrt{841} = 29.

  2. Thus, the maximum displacement xx becomes: xmax=et10(29)+92.x_{max} = e^{\frac{t}{10}}(29) + 92.

  3. Check for the head height:

ightarrow ?$$

  1. Calculate it out to see if xx is less than or equal to 125 to determine if the jumper's head will be above water.

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