How to find height, when given the velocity at that location, starting height, and mass of object? The question:

A roller coaster is 100m above the ground, and weighs 1000kg. At one
  point of the roller-coaster-ride, the velocity is 28 m/s. What is the
  height of the roller-coaster at that point?

All I have so far is the gravitation (potential) energy, however I don't think I need that for this question. 
What I have so far:
$$m = 1000kg$$
  $$h_1 = 100m$$
  $$v_1 = 28m/s$$
  $$g = 9.8 m/s^2$$
  $$E_{P} = 980,000$$
Any help please?
 A: Recall that if the mechanical energy is conserved:
$$K_0+ U_0 = K_f + U_f.$$
In your case you calculated $U_0 = 980,000$ J. Since $K_0 = 0$ you have that:
$$U_0 = \frac{1}{2}mv^2_f + mgh = m\left(\frac{v^2_f}{2}+gh\right) \Rightarrow \frac{U_0}{mg}-\frac{v^2_f}{2g} = h$$
A: You do actually need gravitational potential energy for this problem, and it is central to the concept of this problem, which is energy.  The gravitational potential energy at h=100m is mgh = 980000J, as you calculated.  This total amount of energy will not change (assuming friction does not apply) because in a closed system, mgh+(1/2)mv^2 = C, a constant; in this case, that constant is 980000J.  (That formula just means that the sum of the potential energy and the kinetic energy of a closed system is constant).  If at some point, the velocity is 28 m/s, we get:
(1000kg)*(9.8m/s^2)*h + (1/2)*(1000kg)*(28m/s)^2 = 980000J
Solving for h:
9800*h = 588000;
h = 60

We now know that, assuming the roller coaster's system is closed to external forces, the height at which the coaster is travelling at 28 m/s is 60m.
There are other ways to solve this problem by solving for delta h (in this case 40m) and subtracting that from the original height of 100m to arrive again at 60m, and I encourage you to try other methods yourself now that you know one way to solve this problem.
