The Physics of Pole Vaulting - about the equations: 

In order to determine how high you could theoretically jump, you must know your sprint speed and how tall you are. Again, this assumes that all your energy of motion (kinetic energy) is converted into the "gravitational potential energy" of the jump height.

It turns out that the equation for kinetic energy is expressed as ½ mv², where m equals the mass of the vaulter and v equals the velocity of the vaulter's approach.

Gravitational potential energy is equal to mgh, where m equals the mass of the vaulter, h equals the height of the jump, and g is the acceleration due to gravity on the Earth. Because it depends on gravity, we see right away that if pole vaulting occurred where there was less gravity (such as on the Moon), then the height of a vault would increase proportionately. Indeed, if you could pole vault on an asteroid, its gravity is so weak that you might jump right into space, never to return!

By assuming a complete conversion of kinetic energy into gravitational potential energy, then the two expressions can be set equal to each other. Therefore,

½ mv²=mgh

Note that the mass of the vaulter "m" occurs on both sides. Thus, it "cancels out." We are left with just:

½ v² = gh

Solving for the height, we get:

h=1/2 (v²/g)

This expression needs to be adjusted slightly to account for the fact that the vaulter herself has a height above the ground. This height is different from the height "h" we are trying to figure out. It is based on--but not exactly the same as-- how tall the vaulter is. Rather, it is based on something called "center of mass."

The "center of mass" is defined as the point in a body which acts as if all the body's mass were concentrated there. For an object like a baseball, the center of mass is at the middle of the ball. For a person, the center of mass is roughly half of how tall he or she is. This varies, though, by body type. The program used here makes the reasonable assumption that your center of mass is 55 percent of your height.

Returning to our expression for "h," we have

h= [height of the center of mass] + ½(v²/g)

or

h= 0.55 * [your height] + ½(v²/g)

Because your height is fixed, the only thing that varies is your sprint speed "v."

A measured sprint speed for Dragila is 8.33 m/sec (or 18.7 miles per hour). Her height is 5 feet 8 inches or 1.73 meters. Thus, her center of mass is approximately 55 percent of this, or 0.95 meters. Gravity "g" is a constant at 9.8 m/sec². So the height "h" she should theoretically be able to jump is:

h= 0.95 meters + ½[(8.33 m/sec) x 2/9.8m/sec²]= 0.95 meters + 3.54 meters = 4.49 meters (or about 14 feet 9 inches)

Recall that this is only a first order approximation that only considers sprint speed. Yet as noted by Cliff Frohlich, "the model predicts the height attained remarkably well." In fact, Stacy's most recent record is 15 feet 2 1/4 inches. The difference is accounted for by the fact that the sprint speed used is only an approximation, and no other factors are accounted for.

FURTHER READING:
Frohlich, C., Effect of wind and altitude on record performance in foot races, pole vault, and long jump, American Journal of Physics 53, 726-730, 1984. Reprinted in The Physics of Sports, edited by A. Armenti, American Institute of Physics, New York, 120-124, 1992.