# Jenn Suhr recently won the gold medal in this summer's Olympic games in pole vault. Her record is...

## Question:

Jenn Suhr recently won the gold medal in this summer's Olympic games in pole vault. Her record is 16 ft, 2 in. (She did less than that to win gold this summer.) If all of her gravitational potential energy came only from her kinetic energy, calculate (in m/s) the running speed necessary to make this vault. Assuming the vaulting pole to be elastic, does the pole add energy to this system? Explain.

## Conservation of Energy:

In a closed system, there are two fundamental concepts that need to be understood. If the system is closed, then no energy can flow in our out of the system. Secondly, the energy already inside the system is constant. This means that any actions that cause changes in energy will not change the total energy, but rather only the kinetic and potential energies of the system.

Given:

• {eq}\displaystyle h = 16\ ft\ 2\ in {/eq} is the maximum height

Now let us first convert our height to meters. In 1 foot there are 12 inches so we have our height as:

{eq}\displaystyle h = 16\ ft \left(\frac{12\ in}{1\ ft} \right) + 2\ in {/eq}

{eq}\displaystyle h = 16\ \require{cancel}\cancel{ft} \left(\frac{12\ in}{1\ \require{cancel}\cancel{ft}} \right) + 2\ in {/eq}

{eq}\displaystyle h = 192\ in + 2\ in {/eq}

{eq}\displaystyle h = 194\ in {/eq}

Now we convert this to meters:

{eq}\displaystyle h = 194\ in \left(\frac{1\ m}{39.37\ in} \right) {/eq}

{eq}\displaystyle h = 194\ \require{cancel}\cancel{in} \left(\frac{1\ m}{39.37\ \require{cancel}\cancel{in}} \right) {/eq}

We get:

{eq}\displaystyle h = 4.9276\ m {/eq}

Now the energy that Jenn should have is constant and only varies between kinetic and potential energy. All of her initial kinetic energy should be entirely converted into gravitational potential energy at the maximum height. We thus equate:

{eq}\displaystyle \frac{1}{2} mv^2 = mgh {/eq}

We isolate the speed here:

{eq}\displaystyle v = \sqrt{2gh} {/eq}

We substitute:

{eq}\displaystyle v = \sqrt{(2)(9.8\ m/s^2)(4.9276\ m)} {/eq}

We will get the speed:

{eq}\displaystyle \boxed{v = 9.83\ m/s} {/eq}

Now as for the second question, the pole itself does not add any energy. The pole simply acts as a medium for which the energy transfer takes place. Since the pole is elastic, it stores the energy much like a spring does. This energy allows the storage of the initial kinetic energy, thus converting it to the gravitational potential energy at launch that steadily increases until Jenn reaches maximum height.