Pole Vault!

a) The Law of Conservation of Energy can be written as:

\begin{equation*}
E_{m_i} = E_{m_f}.
\end{equation*}

In this case:

\begin{equation*}
\frac{1}{2} m v_i^2 = mgh.
\end{equation*}

Solving for \(v\), we get:

\begin{equation*}
v = \sqrt{2gh},
\end{equation*}

which, with numerical values, is:

\begin{equation*}
v = 10.87 \, \text{m/s}.
\end{equation*}

b) The velocity is not a function of the mass \(m\). Therefore:

\begin{equation*}
v = 10.87 \, \text{m/s}.
\end{equation*}

c) With the same equation for Conservation of Energy as written in part (a), but solving for \(h\,) we get:

\begin{equation*}
h = \frac{v_i^2}{2g}.
\end{equation*}

Plugging in numerical values yields:

\begin{equation*}
h = 7.97 \, \text{m}.
\end{equation*}

a) In order to find the minimum speed that Thiago needed to have just before using the pole, we should use conservation of energy. Why? Because the energy at the point Thiago starts using the pole must be the same as the energy at the highest point of the jump (they tell us that no energy is lost in the jump). Hence,

\begin{equation}
\label{Olympics_energias}
E_{m_i} = E_{m_f}.
\end{equation}

Now, let’s use a coordinate system located at the floor (see figure 1), so that initially Thiago has no gravitational potential energy (his height \(h\) would be zero, and so \(mgh=0\)):

Figure 1: We place the coordinate system at the ground below Thiago’s initial position.

Since there is no gravitational potential energy, initially the mechanical energy is

\begin{equation}
\label{Olympics_energiaInicial}
E_{m_i} = \frac{1}{2} m v_i^2.
\end{equation}

When Thiago is at the maximum point of the jump, he will have gravitational potential energy (the height is no longer zero). In general, he will be be moving with some speed, and so will have some kinetic energy as well. However, note that they ask us for the minimum initial speed. Clearly, there are many possible (theoretically infinite) initial speeds which would allow Thiago to reach 6.03 meters in height. Very high initial speeds will give Thiago considerable speed at the point of the 6.03 meters mark. Lesser initial speeds will result in lesser speed at the 6.03 meters mark. The minimum initial speed would correspond to a certain speed at which Thiago is barely moving at the point where he reaches the 6.03 meters (if the initial speed was even slower that this minimum one, Thiago would not have enough energy to jump that high). For that minimum speed, we can take Thiago’s speed at the 6.03 meters mark to be approximately zero (specially compared to the initial speed), and so his kinetic energy at that point will also be zero (\(\frac{1}{2}mv_f^2 =0\) because \(v_f=0\)). So the final mechanical energy is solely gravitational potential energy

\begin{equation}
E_{m_f} = mgh.
\end{equation}

Using this result and \eqref{Olympics_energiaInicial} in equation \eqref{Olympics_energias}, we get

\begin{equation}
\label{Olympics_ConservacionEnergia}
\frac{1}{2} m v_i^2 = mgh.
\end{equation}

If we divide by \(m\) everywhere and multiply by 2, we get

\begin{equation}
v^2 = 2gh.
\end{equation}

Finally, take the square root:

\begin{equation}
\label{Olympics_velocidad}
v = \sqrt{2gh}.
\end{equation}

All we need to do is insert the numerical values here:

\begin{equation}
v = \sqrt{2(9.8 \, \text{m/s}^2) (6.03 \, \text{m})}.
\end{equation}

The result is

\begin{equation}
v = 10.87 \, \text{m/s}.
\end{equation}

Note that equation \(v = \sqrt{2gh}\) is also the equation for the speed of an object that is released from a height \(h\).

b) Thiago is 5 kilograms heavier now, how faster does he have to run to make the jump? From equation \eqref{Olympics_velocidad} we see that the initial speed only depends on the height of the jump and \(g\). So, Thiago’s mass is irrelevant! In other words, the minimum speed to perform the jump is exactly the same no matter how massive the person is.

The reader might think this is odd: Would it not be harder for a heavier person to jump a certain height than for a lighter person? Well, it turns out that what matters is the speed at the moment of using the pole, not the mass. Of course, for a heavier person it might be harder to reach a given speed in the same amount of time, after all, the acceleration is inversely proportional to the mass (by Newton’s Second Law). But assuming the heavier and the lighter athletes have the same speed at the moment of starting the jump, they both should be able to reach the same height (assuming, of course, that no energy is lost during the jump).

c) Now, let’s find out the height of a jump in the case the initial speed was that of someone running as fast as Usain Bolt. To do this, we start with equation \eqref{Olympics_ConservacionEnergia}. Let’s write it here again

\begin{equation}
\label{Olympics_ConservacionEnergia2}
\frac{1}{2} m v_i^2 = mgh.
\end{equation}

Now cancel the mass, and divide by \(g\),

\begin{equation}
\frac{v_i^2}{2g} = h.
\end{equation}

If we insert the numerical values

\begin{equation}
\frac{(12.5 \, \text{m/s})^2}{2 (9.8 \, \text{m/s}^2)} = h
\end{equation}

we get

\begin{equation}
7.97 \, \text{m} = h.
\end{equation}

Notice that this is almost two meters higher than Thiago’s record!

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