Capybara!

Using the equations of constant acceleration along Y, try to find the time of the motion. Then, try to find the speed in the horizontal case.

The equation of motion along Y is:

\begin{equation}
y_c=y_{i,c}+v_{iy,c}t-\frac{1}{2}gt^2.
\end{equation}

Using the known variables, and solving for \(t\), we can get:

\begin{equation*}
t=\sqrt{\frac{2H}{g}}.
\end{equation*}

The equation of motion along X, where the motion has constant speed, is:

\begin{equation*}
x_c=v_{ix,c}t.
\end{equation*}

The time was already calculated. Consider that \(x_c=L\) to find the minimum speed for the capybara to reach the other side. Solving for \(v_{ix,c}\), we have:

\begin{equation*}
v_{ix,c}^{\text{min}}=L\sqrt{\frac{g}{2H}}.
\end{equation*}

Using numerical values, we get:

\begin{equation*}
v_{ix,c}^{\text{min}}\approx 4.70\,\text{m/s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

They’ve asked us to find the minimum speed that the capybara needs to reach the other side of the creek. To approach this problem, we will first find the time \(t\) that it takes the capybara to fall a distance \(H=2\,\text{m}\). With this time in mind, we will use the equations of motion along the X axis to find the horizontal distance travelled by the capybara during the fall. By demanding that this distance is larger than \(L=3\,\text{m}\), we can find the minimum speed required for the capybara to successfully and safely travel to the other side.

Let’s find the time that it takes for the capybara to fall. To do that, we start by choosing a coordinate system, as indicated in figure 1.

Figure 1: We place the coordinate system at the bottom of the cliff on the left. The height of this cliff is \(H\), and the horizontal length is \(L\). The horizontal capybara’s initial velocity is also shown.

Note that along Y, the capybara has constant negative gravitational acceleration given the choice of system. Hence, we can write the equation of motion along Y as

\begin{equation}
y_c\,\hat{\textbf{j}}=y_{i,c}\,\hat{\textbf{j}}+v_{iy,c}t\,\hat{\textbf{j}}-\frac{1}{2}gt^2\,\hat{\textbf{j}},
\end{equation}

where \(y_{i,c}=H\) is the initial position along the Y axis of the capybara, \(v_{iy,c}=0\) is the initial speed along the Y axis of the capybara, and \(g=9.8\,\text{m/s}^2\) is the gravitational acceleration. After dropping the vector notation (since everything is along the same axis) and using the fact that the initial vertical speed is zero, we get

\begin{equation}
y_c=y_{i,c}+0-\frac{1}{2}gt^2.
\end{equation}

Because we want to find the time that it takes for the capybara to fall, we set \(y_c=0\) (the final position is zero). Thus, we can write the equation above as

\begin{equation}
0=H-\frac{1}{2}gt^2.
\end{equation}

Solving for time \(t\), we get

\begin{equation}
\frac{1}{2}gt^2=H,
\end{equation}

which, after multiplying by \(\frac{2}{g}\) on both sides, becomes

\begin{equation}
t^2=\frac{2H}{g}.
\end{equation}

After taking the square-root on both sides, this yields

\begin{equation}
\label{time}
t=\sqrt{\frac{2H}{g}}.
\end{equation}

Hence, we’ve found an expression for the time of the fall. We can now use this time in the equations along X to find the minimum horizontal speed.

Along X, the capybara has constant velocity (since it follows a semi-parabolic motion). Given the system, the velocity along X is positive, and so we get

\begin{equation}
x_c\,\hat{\textbf{i}}=x_{i,c}\,\hat{\textbf{i}}+v_{ix,c}t\,\hat{\textbf{i}},
\end{equation}

where \(x_{i,c}=0\) is the initial position of the capybara and \(v_{ix,c}\) is the initial speed of the capybara along the X axis. After dropping the vector notation to focus on the magnitudes and using the fact that \(x_{i,c}=0\), this becomes

\begin{equation}
x_c=v_{ix,c}t.
\end{equation}

In order for the capybara to reach the other side safely, we must demand that \(x_c>L\) (the final horizontal position must be greater than distance \(L\)). We can then use the expression for \(x_c\) given above to write

\begin{equation}
v_{ix,c}t>L.
\end{equation}

Using the expression for the time of fall given by equation \eqref{time} in this inequality, we get

\begin{equation}
v_{ix,c}\left(\sqrt{\frac{2H}{g}}\right)>L.
\end{equation}

Solving for \(v_{ix,c}\), we obtain

\begin{equation}
v_{ix,c}>L\sqrt{\frac{g}{2H}}.
\end{equation}

Thus, if the velocity \(v_{ix,c}\) is larger than \(L\sqrt{\frac{g}{2H}}\), the capybara will reach the other side safely. Thus, the minimum velocity that the capybara must have is precisely \(L\sqrt{\frac{g}{2H}}\), that is,

\begin{equation}
v_{ix,c}^{\text{min}}=L\sqrt{\frac{g}{2H}}.
\end{equation}

Using the numerical values, we get

\begin{equation}
v_{ix,c}^{\text{min}}=(3\,\text{m})\sqrt{\frac{9.8\,\text{m/s}^2}{2(2\,\text{m})}},
\end{equation}

which is

\begin{equation}
v_{ix,c}^{\text{min}}\approx 4.70\,\text{m/s}.
\end{equation}

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