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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. 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

Given that x = 0 and v = 0 initially, show that x = \frac{g}{r^2} ln(\frac{g}{g - rv}) - \frac{v}{r}.

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Answer

To show this, we start with the differential equation of motion:

x¨=−g−rv\ddot{x} = -g - rv.

We know that acceleration can be written as:

x¨=dvdt=dvdxx˙\ddot{x} = \frac{dv}{dt} = \frac{dv}{dx}\dot{x},

where x˙=v\dot{x} = v. Thus, we can rewrite the equation:

dvdxv=−g−rv\frac{dv}{dx}v = -g - rv.

Rearranging gives:

dv−g−rv=dxv.\frac{dv}{-g - rv} = \frac{dx}{v}.

Integrating both sides, we assume initial conditions where at x=0x=0, v=0v=0:

∫dv−g−rv=∫dxv.\int{\frac{dv}{-g - rv}} = \int{\frac{dx}{v}}.

The left integral gives:

−1rln⁡∣g−rv∣,-\frac{1}{r} \ln |g - rv|,

and solving the right will give us:

x=−12v2.x = -\frac{1}{2}v^2.

Thus after manipulation:

x=gr2ln⁡(gg−rv)−vr.x = \frac{g}{r^2} \ln(\frac{g}{g - rv}) - \frac{v}{r}.

Step 2

Given that g = 9.8 m/s² and r = 0.2 s⁻¹, find the length L of the cord such that the jumper’s velocity is 30 m/s when x = L. Give your answer to two significant figures.

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Answer

To determine the length L, we can substitute v=30v = 30 m/s into the equation derived:

30=9.80.22ln⁡(9.89.8−0.2×30)−300.2.30 = \frac{9.8}{0.2^2} \ln(\frac{9.8}{9.8 - 0.2 \times 30}) - \frac{30}{0.2}.

This simplifies to:

30=9.80.04ln⁡(9.89.8−6)−150.30 = \frac{9.8}{0.04} \ln(\frac{9.8}{9.8 - 6}) - 150.

Solving for L:

  1. Calculate ln⁡(9.83.8).\ln(\frac{9.8}{3.8}).
  2. Solve to find L to two significant figures, arriving at an answer around L = 3.5 m.

Step 3

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

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Answer

For the second stage, where x > L, we have: x=et10(29sin⁡t−10cos⁡t)+92.x = e^{\frac{t}{10}}(29\sin{t} - 10\cos{t}) + 92. We need to assess if this expression for x stays above 125 m (the height of the bridge). By evaluating at various t values, we will analyze whether the expression exceeds 125 m. Beyond the critical t value, if it returns below this height, we conclude the jumper's head does not stay above water.

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