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.
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!
a complete conversion of kinetic energy into gravitational potential
energy, then the two expressions can be set equal to each other.
the mass of the vaulter "m" occurs on both sides. Thus, it "cancels
out." We are left with just:
½ v2 =
the height, we get:
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."
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.
to our expression for "h," we have
of the center of mass] + ½(v2/g)
* [your height] + ½(v2/g)
height is fixed, the only thing that varies is your sprint speed "v."
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:
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
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.
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.