Originally Posted by **arbolis** Could you provide the work of your professor? So that we can maybe help you to understand what he did. |

I don't have a scanner but I'll try to explain it as clearly as I can here:

The final formula is t'=t(1-sqrt(1-f)) where t' is the "last 1 second of the fall", t is the total time, f is h'/h (h'/h = .7212 in this problem) where h' is the distance traveled in the last 1 second and h is the total distance.

I understand most of his reasoning except for one step.

Here are his steps:

EQUATION ONE

1) $\displaystyle h=\frac{gt^2}{2}$

2) $\displaystyle t=\sqrt {\frac{2h}{g}}$

EQUATION TWO

3) $\displaystyle h-h'=\frac{g(t-t')^2}{2}$

4) This is where I'm confused. Somehow he used the two above equations to get the equation:

5) $\displaystyle (1-f)=\frac{(t-t')^2}{t^2}$

6) $\displaystyle (1-f)=(1-\frac{t'}{t})^2$

7) $\displaystyle 1-\frac{t'}{t}=\sqrt {1-f}$

8) $\displaystyle t'=t(1- \sqrt {1-f})$

By the way, I figured out how to plug in the correct constants into his equation to get the correct time. I used the time to find the distance. However, I would still like to know how he derived his equation.