Question

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging river and attach a 31-m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 42 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

Answer 1

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

$V_2^2=V_o^2-2*g*\Delta h$

$V_2^2=-2*10*(-31)$

$V_2=24.9m/s$   This will be our initial speed for a balance of energy.

By conservation of energy:

$m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0$

Where

$h$ is your height at its maximum elongation

$h_o$ is the height of the bridge

$l_o$ is the length of the unstretched bungee cord

$800h+21*(79-h)^2-63200-24800.4=0$

$21h^2-2518h+43060.6=0$ Solving for h:

$h_1=20.66m$  and $h_2=99.24m$  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

Answer 2

Answer: using the conservation of potential energy stored in spring giving that at maximum amplitude velocity becomes zero.

Mgd= 1/2k(d-l)^2..... equation 1

M= 80kg=mass , g= 10m/s^2 =gravity, d=?=length of fully extended bungee rope, l=31m= length of bungee rope before extension, k=42N/m= spring constant

Simplifying equation above gives

2Mgh/k= d^2 - 2dl + l^2 ....eq 2

Substituting figures into the equ above gives

0 = d^2 - 100.1d +961 ...equ 3

Equ3 can be solved since it is a quadratic equation

d= (-b +or- square root (b^2 - 4ac))/2a ....equa4

Where a=1, b= -100.1, c= 961

Substituting figures into eequa4

d= 89.34m

So therefore the height above the river to me when bungee is fully extended is= 110 - 89.34

= 20.66

Explanation: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.

Assuming that it performs simple harmonic motion.