Bungee Jump (Modelling) 
A bungee jumper of mass $m$ attaches a light elastic string to his foot, and the other end to railing of a bridge on which he stands. The string has natural length $l_0$ and modulus of elasticity $\lambda$. Write down the energy equation. 

(i)To make life a little easier for the calculations I'm going to choose the         origin to be a distance $h$ below the point of suspension.
I understand the principles of energy conversion during a bungee jump. So during the first interval, stepping of the bridge, the only force acting on the system(bungee jumper) is gravity(free fall). As the bungee jumper falls, the gravitational potential energy is converted into kinetic energy. When he reaches the point where the bungee cord begins to stretch, gravitational potential energy begins to be converted into the elastic potential energy of the cord. Eventually, all of the kinetic energy is also converted into potential energy of the cord, and he comes to rest for a short period of time. 
Letting $h$ be the distance the jumper falls I have 
$$PE_i=mgy=mgh$$
$$PE_f = mgy = mg(0)= 0$$
The potential energy of the elastic string is $\frac{1}{2}kx^2$ where $x$ is the amount the string has streched. At the bottom of the fall the string is stretched by an amount $(h-l_0)$. But the cord doesn't begin to stretch until he falls a distance $l_0$. Therefore the final elastic potential energy energy is $$\frac{1}{2}kx^2 = \frac{1}{2}(h-l_0)^2 = \frac{\lambda}{2l_0}(h-l_0)^2$$
Since the relationship between the string constant and the elasticity is $$k=\frac{\lambda}{l_0}$$.
Thus the energy equation I found to be $$mgh = \frac{\lambda}{2l_0}(h-l_0)^2$$

Use conservation of energy to find the maximum distance the jumper descends after stepping off the bridge.

The maximum distance corresponds to the lowest point of the fall where the velocity is 0 (momentarily).When he is standing on the top, he is at rest. Therefore his kinetic energy is zero. 
When he is at the bottom, he is also at rest so the kinetic energy is also 0 (since the velocity is zero.)
That being said, we need to be careful because there is also potential energy in the string as it stretches.
I think I need to solve this equation $$mgh = \frac{\lambda}{2l_0}(h-l_0)^2$$ but I don't know how to do it. Is this a differential equation?
I would really appreciate the help :)
 A: Your "energy equation" is a correct description of what happens at the
lowest point that the jumper reaches,
given your choice of origin for the coordinates and your definition of
the constant $h.$
(I call $h$ a constant because the alternative is that the height of the bridge above your origin of coordinates is variable, which does not make much sense.)
The parameters $m,$ $g,$ $\ell_0,$ and $\lambda$ also are constants,
so each side of the equation is a constant expression.
As long as the requested "energy equation" only needs to describe
the energy at one point in the jump by equating two constant expressions,
then it's OK. (If the author of the question has been showing "energy equations" that don't fit that description, you may want to think about this more.)
My result for the distance to the lowest point of the jump agrees with yours, but I do not think you really "make life a little easier for the calculations" by choosing the origin to be the lowest point the jumper reaches.
You do ensure that the $y$ coordinate and potential energy both will be non-negative throughout the jumper's motion, but negative numbers do not always make calculations more difficult.
In orbital mechanics, for example, I think the best way to represent potential energy is as a quantity that is always negative.
I can think of at least three other obvious origins for the vertical axis. One is at the height of the bridge, which in some ways is an easier place to put the origin than your choice (for example, you can immediately write a formula for potential energy that doesn't involve any unknown constants), but I think it is still not the easiest origin.
A second possible origin is the point at which the bungee first
goes taut, a distance $\ell_0$ below the bridge.
If $z$ is height above that point, the jumper starts at $z = \ell_0$
and the lowest point the jumper reaches is $z = -h$
(using your value of $h$), and the equation of energy is
$$\frac12 mv^2 + \frac{\lambda}{2l_0}z^2 + mgz = mg\ell_0$$
whenever the bungee is under tension.
At the lowest point of the jump ($v=0$), this equation becomes
$$\frac{\lambda}{2l_0}z_0^2 + mgz_0 = mg\ell_0,$$
where $z_0$ is the distance below $z=0$ at the bottom of the jump,
and the maximum distance the jumper descends is $\ell_0 + z_0.$
This still has you solving a quadratic equation, but you don't have to
expand a term like $(h - \ell_0)^2$ in order to set up the quadratic,
so I think it's significantly easier to work with.
A third possible origin is the point at which the jumper is in static equilibrium, that is, when the bungee is stretched just enough so that its tension equals the jumper's weight.
This occurs at a distance $$\ell_0 + \frac{mg}{\lambda}\ell_0$$
below the bridge.
You have to be a bit careful in finding the potential energy at that point, because although the net force on the jumper is $-mg$ for the first $\ell_0$ distance fallen, the net force then decreases linearly to zero over the distance $\frac{mg}{\lambda}\ell_0$ from where the bungee goes taut to the point of equilibrium, and the average force over that distance is just
$-\frac12 mg.$
Setting the potential energy at the equilibrium point to zero,
that is, $PE(0) = 0,$
the jumper's potential energy just before jumping off the bridge was
$$
PE_i = mg\ell_0 + \frac12 mg\left(\frac{mg}{\lambda}\ell_0 \right)
= \left(1 + \frac{mg}{2\lambda} \right) mg\ell_0.
$$
The spring constant is the same between the equilibrium point and the bottom
as it is between the point where the bungee goes taut and the bottom,
and the total vertical force on the jumper increases linearly from zero
at that same rate as the jumper falls below the equilibrium point.
In other words, using the equilibrium point as our origin, from the
equilibrium point to the bottom the total force on the jumper follows
Hooke's Law with the same spring constant as before;
that accounts for all the change in potential energy over that distance,
including gravitational potential, so we don't need to account for gravity separately.
The jumper's potential energy at any vertical location $u$ below the equilibrium point (that is, at $u < 0$) is
$$
PE(u) = \frac12 ku^2 = \frac{\lambda}{2\ell_0} u^2,
$$
and the equation of energy at any such point is
$$
\frac12 mv^2 + \frac{\lambda}{2\ell_0} u^2 
= \left(1 + \frac{mg}{2\lambda} \right) mg\ell_0.
$$
Setting $u_0$ to be the (positive) distance between the equilibrium point and the lowest point of the jump, then at $u = -u_0$ (the bottom of the jump), where $v = 0,$ the energy equation simplifies to
$$
\frac{\lambda}{2\ell_0} u_0^2
 = \left(1 + \frac{mg}{2\lambda} \right) mg\ell_0.
$$
This is by far the easiest among all the quadratic equations we've seen yet
for this question: just multiply both sides to get $u_0^2$ all by itself on the left, and then take the square root of both sides, with a $\pm$ sign.
The correct solution is the positive square root, of course,
and the maximum distance the jumper falls is
$$\ell_0 + \frac{mg}{\lambda}\ell_0 + u_0.$$
After solving for $u_0$ you can see that this is the same distance you found
(taking the $+$ option of the $\pm$ sign).
A: Looks like straightforward algebra from there.  Treat h as your variable and everything else as a constant.
You can distribute the right hand side of the equation, collect your h terms on one side, and then solve for h.
A: $$\begin{align}
    mgh &= \frac{\lambda}{2l_0}(h-l_0)^2 \\
    mgh &= \frac{\lambda}{2l_0}(h^2-2hl_0+l_0^2) \\
    mgh &= \frac{\lambda}{2l_0}h^2 - \lambda h +\frac{\lambda l_0}{2} \\
\end{align}
$$
Then I found this quadratic equation that needs to be solved using the quadratic formula
$$
\begin{align}
  \frac{\lambda}{2l_0}h^2 - h(mg+\lambda) + \frac{\lambda l_0}{2} = 0
\end{align}
$$
Is this correct so far? I ended up with 
$$ h = \frac{(mg+\lambda)l_0\pm l_0\sqrt{mg(mg+2\lambda)}}{\lambda} $$
This looks a little messy though so I'm not sure it's correct. 
