Flick Football!

Find the time that it takes for an object in free fall motion to reach the ground. Then, find each horizontal distance with the equation of motion for an object with constant speed.

Since the almond follows a ‘semiparabolic’ motion because it is hit horizontally, we know that the horizontal speed is constant. Hence, the horizontal distance (the distance in X) is given by:

\begin{equation*}
d_x=v_x t.
\end{equation*}

Consider the motion along the Y axis. Vertically, the almond has constant acceleration, and in general, the equation of motion for an object with constant acceleration is:

\begin{equation*}
\vec{y}_f =  \frac{1}{2} \vec{g} t^2 + \vec{v}_{i_y}t + \vec{y}_i.
\end{equation*}

With the given values, and solving for \(t\), we get:

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

To find the distance difference, we can write:

\begin{equation*}
\Delta D = |d_{x_1} – d_{x_2}| = \left|v_x t – v_x t \right| = \left|v_x \sqrt{\frac{2 y_1}{g}} – v_x \sqrt{\frac{2 y_2}{g}} \right|,
\end{equation*}

where \(d_{x_1}\) and \(d_{x_2}\) are related with the time variable \(t\) found before, which is also a function of \(y_1\) and \(y_2\) respectively (the given heights).

Using the numerical values we can get:

\begin{equation*}
\Delta D = 0.2 \, \text{m}.
\end{equation*}

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

We need to find the difference between the horizontal distance traveled by the almond when flicked at an initial height of 1 m and the horizontal distance that it travels when flicked at a height of 1.5 m (in both cases, the almond is flicked horizontally). So, we have to find an expression that expresses the horizontal distance traveled by the almond given a certain initial height. Once we find such an expression, we can just put in the numbers. The key to finding that expression rests in noting that the almond follows a semi-parabolic type of motion.

Let’s start by choosing a coordinate system (figure 1). We’ll place one on the ground, directly below the point at which the almond is flicked and where the X axis points in the almond’s direction of motion.

Figure 1: We place the coordinate system on the floor just below the almond’s initial position. The X axis is directed to the left along the almond’s trajectory.

Now, the total horizontal distance for any kind of projectile depends only on the horizontal velocity and the time. This is true no matter what the initial angle is or the initial height or the initial speed of the projectile. Since the almond follows a parabolic motion (or ‘semiparabolic’ motion, since it’s thrown horizontally), we know that the horizontal speed is constant. Hence, the horizontal distance (the distance on X) is given by

\begin{equation}
\label{Flick_dvxt}
d_x=v_x t,
\end{equation}

where \(v_x\) is the velocity along X, and \(t\) is the time that the almond is flying. In this case \(v_x\) is just the initial speed \(v_i\), since the initial angle is zero, and so the only component for the initial velocity is the X component (the almond is thrown completely horizontally). So we can rewrite equation \eqref{Flick_dvxt} as

\begin{equation}
\label{Flick_dvt}
d_x=v_i t.
\end{equation}

Now, let’s compute the total time that the almond is flying. To do this, let’s consider the motion along the Y axis. Vertically, the almond has constant acceleration, and in general, the equation of motion for an object with constant acceleration is

\begin{equation}
\label{Flick_yfVector}
\vec{y}_f =  \frac{1}{2} \vec{g} t^2 + \vec{v}_{i_y}t + \vec{y}_i.
\end{equation}

According to our coordinate system, the initial position (height) is positive, the gravitational acceleration is negative, there is no initial velocity in Y (again, the almond is flicked horizontally), and the final position is also zero (because we’re only interested in the distance that the almond travels until it touches the floor). Hence, equation \eqref{Flick_yfVector} becomes

\begin{equation}
\label{Flick_yf}
0 \, \hat{\textbf{j}} = – \frac{1}{2} g t^2 \, \hat{\textbf{j}} + y_i \, \hat{\textbf{j}}.
\end{equation}

Given this equation, we can easily solve for the time \(t\). First, let’s move the term \(- \frac{1}{2} g t^2\) to the left and focus on the magnitudes:

\begin{equation}
\frac{1}{2} g t^2 = y_i.
\end{equation}

Then, multiply both sides by \( \frac{2}{g} \) and take the square root to get

\begin{equation}
\label{Flick_tiempo}
t = \sqrt{\frac{2 y_i}{g}}.
\end{equation}

Notice that the time of the almond’s flight depends on the initial height but not on the initial speed! So this equation works even for the case where the initial speed is zero. In other words, equation \eqref{Flick_tiempo} shows that if an object is released (with no initial speed) from a certain height, or if it is thrown very rapidly in a horizontal direction, the time that it takes to reach the floor is exactly the same. Of course, if it’s not thrown horizontally, the total time before reaching the floor will also depend on the initial velocity along Y.

If we use this time in equation \eqref{Flick_dvt}, we get

\begin{equation}
d_x = v_x \left( \sqrt{\frac{2 y_i}{g}} \right).
\end{equation}

Notice that this expression gives us the total distance as a function of the initial height and the initial speed. Obviously, the higher the initial height, the longer the distance travelled by the almond. So, all we need to do now is find the difference between the distance corresponding to the two initial heights. If we call one height \(y_{1}\) and the other one \(y_{2}\), we get

\begin{equation}
\Delta D = |d_{x_1} – d_{x_2}| = \left|v_x \sqrt{\frac{2 y_1}{g}} – v_x \sqrt{\frac{2 y_2}{g}} \right|,
\end{equation}

where the vertical bars \(|\) indicate that we only care about the absolute values or magnitudes of the terms (the distance is always a positive number).

Let’s factorize the initial speed \(v_x\) (we can do this because it is positive):

\begin{equation}
\Delta D = v_x \left| \sqrt{\frac{2 y_1}{g}} – \sqrt{\frac{2 y_2}{g}} \right|
\end{equation}

Finally, let’s insert the numerical values

\begin{equation}
\Delta D = (2\, \text{m/s}) \left| \sqrt{\frac{2 (1 \, \text{m} ) }{(9.8 \, \text{m/s}^2)}} – \sqrt{\frac{2 (1.5 \, \text{m})}{(9.8 \, \text{m/s}^2)}} \right|
\end{equation}

to get

\begin{equation}
\Delta D = 0.2 \, \text{m}.
\end{equation}

Of course, if we had subtracted \(d_{x_1}\) from \(d_{x_2}\) instead, we would have obtained the exact same result because the absolute value of \((a-b)\) is the same as the absolute value of \((b-a)\) (this is why we used the absolute value in the first place).

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