Finding height with constant acceleration

[!!!]

Homework Statement

What's the cliff's height (m) if a fallling boulder takes 1.3 s to fall the last 3rd of the way to the ground?
Air resistance is ignored, and the problem requires a quadratic formula.

Homework Equations

I think it's the constant x-acceleration formulas, which I can't type here because of the super subscripts/ subscripts.

The Attempt at a Solution

Actually, I don't even know how to begin this or set this up. =/ How do I solve this equation and determine which one of the given answers is correct? The answers in my textbook says it's either 246 m and 2.51 m, but how do I solve to get them?

 

[cepheid]

Call the cliff's height h. Let y(t) be the distance above the ground as a function of time.

y(t) = h - \frac{1}{2}gt^2

I'm assuming it starts from rest. Solving for t, the total time to fall is:

t = \sqrt{\frac{2h}{g}}

1.3 s before that elapsed time, the ball has fallen 2/3 of the way:

y\left(\sqrt{\frac{2h}{g}} - 1.3 s\right) = \frac{h}{3} = h - \frac{1}{2}g\left(\sqrt{\frac{2h}{g}} - 1.3 s\right)^2

Now I think you can just solve for h. Let me see if I get one of those answers.

 

[cepheid]

edit: Yes, I do get both of those answers.