Question

A pole vaulter has a frontal area of 1.32m2, a coefficient of drag of 0.72, and a body mass of 82.4kg. As he approaches the vault, he is running at a speed of 8.4m/s into a 1.34 m/s headwind. His center of mass is at a height of 1.05m. Assuming all of his kinetic energy is transferred to gravitational PE, how high will his center of mass be at the apex of his vault?

Answer 1

To express the basic physics in mathematical terms, we must equate the kinetic energy of the vaulter to the gravitational potential energy corresponding to the point of maximum height.

Therefore,
(1/2)MV² = Mg(Δh)

Where:

M is the mass of the pole vaulter
V is the velocity of the pole vaulter just before the vault
g is the gravitational constant, equal to 9.8 m/s²
Δh is the change in height of the pole vaulter's center of mass

From the above equation,

Δh = (1/2)V²/g (the mass M cancels out)

Δh= 0.5*(8.4^2)/9.8= 3.6m

Since his center of mass is 1.05 m off the ground, this means that his vault height (off the ground) is 4.65 m.