Bungee jumping

Let state that $E_{p} = mgh$ is potential energy at the top relative to the bottom of the bridge and $E_{p} = \frac{1}{2}kx^2$ is elastic potential energy of cord at the bottom extremity of the bridge (remember that $h = L + x$ where $L$ denotes the length of bungee cord and $x$ denotes the stretch of bungee cord). Thus we can write that:

$mgh = \frac{1}{2}kx^2$

$80 \times 9.8 \times (3 + x) = \frac{1}{2} \times 100 \times x^2$

Solve the equation as we get $x = 18.256$ m hence $h = 18.256 + 3 = 21.256$ m

I used the law of conservation of energy.

Let the minimum height of the bridge be $h$ .

Notice that on top of the bridge, the gravitational potential energy of the jumper is equal to $mgh$ , but both his kinetic energy & the elastic potential energy of the cord is zero.

When the jumper reaches the bottom of the bridge, his gravitational potential energy is zero, so is his kinetic energy. But the elastic potential energy of the cord is equal to $\frac{1}{2}kx^2$ where $k$ is the co-efficient of elasticity & $x$ is the amount of stretch.

It is quite clear that $x=(h-3)$ .

So, $mgh=\frac{1}{2}k(h-3)^2$ .

And we have to solve for $h$ .

Plugging in the values and arranging the terms gives us the following quadratic equation:

$25h^2-542h+225=0$ .

Solving this gives us two values of $h$ . One of them is smaller than $3$ meters, so we'll disregard that one.

The other one is our desired result $h\approx 21.276$ meters.

Suppose the drop to the ground below is a distance $h$ . Then the jumper begins with potential energy $mgh$ . When the cord stretches all the way to the ground, it will have stretched $h-3$ meters from its resting position. At that point, all of the energy will be stored in the bungee cord, $E = \frac{1}{2}k x^2 = 50(h-3)^2$ . By the conservation of energy, we have $(80)(9.8)h = 50(h-3)^2$ which has sensible solution $h=21.276$ .

Given the above, we simply use conservation of energy in the form $E_g=E_e$ , giving us: $mg(h+3)=\frac{1}{2}\kappa h^2$ Solving for $h$ gives us two results, we take the positive, $h \approx 18.2566\text{ m}$ and, since our starting position is 3 metres above the bridge, we have that: $h_f=h+3\approx 21.2566\text{ m}$

$m g (h+3)=\frac{1}{2} k h^2\\ h=18.2566$ Add back 3m because the bridge is 3m higher than the unstretched height of the spring.

This solution can be done using the Conservation of energy technique, where there are no external Force working . Use : The initial Energy is equal to the final Energy ==> Ei = Ef. What is the initial energy? The initial energy is the potential energy where the jumper stands. But firstly, we assume that the initial bungee Cord is L and The final bungee cord is L+x where x is a constant (additional long of chord). Let's do this ! Ei = Ef ==> m . g . (L+x) = 1/2 . k . x^2 (We can also assume that the final Energy is fully converted into form of Bungee cord elastic Energy , since the bungee jumper doesn't have any velocity left and we assume that **the final position of the bungee is 0 ). Finishing the sequence, we get k.x^2 - 2.m.g.x - 2.m.g.l = 0.Hence we get x = ( 2.m.g +V [ 4.m^2.g^2 + 8.k.m.g.L ] ) /2k..In the simple form, we get x = ( m.g + V [ m.g { m.g + 2.k.L } ] ) / k.. Substitute m = 80 kg, g = -9.8 m/s^2, k =100 N/m and L = 3 m. From the calculation, we get x is approximately 18.25 m. That's the maximum for the chord to stretch. 7. So, the minimum height of bridge must be 18.276 + L = 18.276 + 3 = 21.276.. (Remember that the 18.276 is just the x, where x is the additional because of the elasticity. The total is actually L + x).

Potential energy at the top of the bridge: $mgh$

Potential energy due to the cord at zero level: $k(h-3)^2/2$

The minimum safe height implies that the jumper stops moving down at zero level exactly i.e. has zero kinetic energy, so we can equate these two terms. Solving for $h$ yields $h = 21.276 \text{ m}$ .

Consider the jumpers energy at the start, which is all potential and therefore of magnitude $mg(3+h)$ . This energy is the same throughout while the string is slack, so for the jumper to reach rest at the bottom of the bridge, the bungee cord must take away all of the jumpers energy.

Gives us: $mg(3+h)$ = $\frac {1}{2}$ $k$ $h^{2}$

Rearrange for a quadratic in h: $h^{2}$ - ( $\frac {2mg}{100}$ ) $h$ - $\frac {6mg}{100}$ = 0

Sub values in and use quadratic formula to get 18.276. Add 3 metres to the answer (length of string as it stretched) to get 21.276

Just conservation of energy, as the man get down, the height that he will get is 3+x where x is a variable, rather saying, x will be the elongation of the spring, so lets do the counts:

$\frac{kx^2}{2}=mgh$

$\frac{kx^2}{2}=mg(3+x)$

$50x^2=80.9,8(3+x)$

$5x^2-78,4x-235,2=0$

Solving this, x will be 18,276m, the height asked is h=3+18,276. Finally, h=21,276m