The AJ Hackett Macau Tower Bungee Jump is 233m high, making this jump a Guinness World Record holder for the Highest Commercial Bungee Jump in the world. And in this report, we are going to investigate the velocity and the displacement of a person with mass $m$ kg during the jump.
For safety concerns, the minimum height is set to be 70m above the ground.
We shall divide the jump into two parts: Before the cord is stretched, and after the cord is stretched.
Let us assume that the cord is \( L \) meters long, which means that the cord will not exert any forces on the jumper until he is \( L \) meters below the platform. Then, according to Newton's Second Law, we know that \begin{equation}F_{\textrm{net}}=ma\end{equation} where \( m \) is the mass (in \( kg \) ) and \( a \) is the acceleration (in \(ms^{-2}\)).
For simplicity sake, we assume that there would be no air resistance throughout the entire bungee jump and we take the downward direction as the positive direction.
Before we begin any computations, let us define some of the meaningful variables: \begin{align*} h &= \textrm{ the height of the jumper above the ground, in meters} \\ t &=\textrm{ the time, in seconds, elapsed since the jump} \\ s &=\textrm{ the distance travelled by the jumper, in meters} \end{align*} So automatically \(h_0=233 \) and \( t_0=0 \).
Since the air resistance is negligable, the only force acting on the jumper is his weight. Hence, \(F_{\textrm{net}}=ma=mg\), where \( g\approx 9.81 ms^{-2} \) is the gravitational constant.
Since the only acting force is the jumper's weight, it is safe to assume that the acceleration of the jumper is constant.
Hence, by our high school physics, we know that \begin{equation*}s=\dfrac{1}{2}gt^2 \end{equation*} and the velocity \( v \) is given by \begin{equation*}v=gt.\end{equation*}
Obviously, if \( s=L \), then \( s=\dfrac{1}{2}g{\tau}^2 \), where \( \tau \) is the time required for the jumper to travel \( L \) meters in the jump. By simple change of variables, we have \begin{equation*} \tau = \sqrt{\dfrac{2L}{g}}\end{equation*} and the corresponding velocity is given by \begin{equation*} v=\sqrt{2Lg}\end{equation*}
After the jumper falls \( L \) meters below the platform, the cord is being stretched past its natural position and a second force acts on the jumper: the stretching force of the cord \( F_c \). By Hooke's Law, the formula is given by \begin{equation*} F_c=-kx \end{equation*} where \begin{align*} k &= \textrm{ spring constant of the cord, and} \\ x &=\textrm{ the distance which the cord stretched past its natural position} \end{align*} So in our context, we have \begin{equation} F_c(h) = \begin{cases} -kx, & \textrm{if \( h\leq 233 - L\)} \\ 0, & \textrm{if \( h > 233-L \)} \end{cases} \end{equation}
Hence, combining the equations, we have \begin{equation} F_{\textrm{net}}=ma=mg + F_c(h) \end{equation}
By rewriting\( a=x^{\prime \prime} \) (as the second derivative of the stretching distance), and by considering \( x \) as a function of \( T=t-\tau \), which represents time elapsed once the cord begins to stretch, we have
\begin{equation} m x^{\prime \prime} = mg - kx \end{equation}By rearranging terms, we get a nonhomogeneous second order differential equation \begin{equation} m x^{\prime \prime} + kx = mg \end{equation}
Let \( x(T)=A\cos(w_0 T) + B\sin(w_0 T)+\dfrac{mg}{k}\), where \(A, B\) are some constants and \begin{equation*} (w_0)^2 = \dfrac{k}{m}. \end{equation*} In the following, we shall prove that the function \( x(T)\) is indeed the solution of the equation \begin{equation*} m x^{\prime \prime} + kx = mg. \end{equation*}
First of all, notice that \( x^{\prime } =-A{w_0}\sin({w_0}T)+B{w_0}\cos({w_0}T) \) and \( x^{\prime \prime}=-A{w_0}^2\cos({w_0}T)-B{w_0}^2\sin({w_0}T).\) Then, by substituting \(x^{\prime \prime}\) and \(x\) into our ODE, we have
\begin{equation*} mx^{\prime \prime}+kx = \cos({w_0}T)(Ak-Am{w_0}^2) + \sin({w_0}T)(Bk-Bm{w_0}^2) + mg = mg . \end{equation*}Hence, we know that \( x(T)=A\cos(w_0 T) + B\sin(w_0 T)+\dfrac{mg}{k} \) is indeed the solution of our ODE.
Notice that by our previous results, we know that \( x(0)=0, x^{\prime}(0)=\sqrt{2Lg}\). So by these two equations, we get \begin{equation*} A=-\dfrac{mg}{k}\end{equation*} and \begin{equation*} B=\sqrt{\dfrac{2mLg}{k}}.\end{equation*}
Since \( x(T) \) is composed of a sine function and a cosine function, this function describes a periodic, or simple harmonic, motion of the jumper. And the period of this function is given by \( \dfrac{2\pi}{w_0}\).
In addition, the maximum streching distance \( x_{\max} \) is given by \begin{equation*} x_{\max}=A\cos(p) + B\sin(p) - A \end{equation*} where \begin{equation*} p=\tan^{-1}\left(\dfrac{B}{A}\right)=\tan^{-1}\left(-\sqrt{\dfrac{2Lk}{mg}}\right). \end{equation*}
Using the above information, we know that the height of the jumper is given by \begin{equation*}h(t) = \begin{cases} 233 - \dfrac{1}{2}gt^2,& \textrm{if \(t\leq \tau\)} \\ 233 - L - \left(A\cos(w_0 (t-\tau)) + B\sin(w_0 t-\tau)+\dfrac{mg}{k} \right),& \textrm{if \( t>\tau \) } \end{cases} \end{equation*} where \( \tau=\sqrt{\dfrac{2L}{g}}\).
Below is the graph of \( h(t)\) against \(t\):
The above diagram is an interactive diagram in which you may adjust the values of \( L, k \) and \( m \). Suggest a tuple \((L, k, m)\) which satisfies the requirement of the AJ Hackett Macau Tower Bungee Jump, and answer the following questions:
When does the cord begin to stretch?
What is the velocity of the jumper when the cord begins to stretch?
What is the value of \( t \) when the jumper reaches its minimum height?
What is the minimum height of the jumper?