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Physics of Media Arts: Conceptual Physics for Students of the Media Arts

Aaron Wong

Section 5.2 Projectile Motion

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Projectile motion is the name of the behavior of objects in motion under the influence of a constant gravitational acceleration. The mechanism that starts everything into motion can be anything from a gun, to a baseball bat hitting a ball, to someone or something dropping an object. For projectile motion, we only care about what happens after the object is no longer under the influence of whatever it is that caused it to start moving.

The Projectile Motion Formulas.

The formulas for projectile motion are an application of the constant acceleration formulas we had in the previous section. In the vertical direction, the constant acceleration is \(a_y = -g\text{,}\) where \(g\) is the acceleration due to gravity. On earth, this is approximately \(9.8 \text{m}/\text{s}^2\text{.}\) In the horizontal direction, there is no acceleration, so \(a_x = 0\text{.}\) Notice that this means that we are ignoring the effects of air resistance. When put together, this gives us the following pair of equations for projectile motion:
\begin{align*} x(t) \amp = v_{0x}t + x_0 \\ y(t) \amp = -\frac{gt^2}{2} + v_{0y}t + y_0 \end{align*}
It may take some time to get used to the notation, but this is a standard way of writing it. This is what the symbols mean:
  • \(v_{0x} = v_x(0)\text{:}\) This is the initial velocity in the horizontal direction.
  • \(x_0 = x(0)\text{:}\) This is the initial horizontal position of the object. Typically, the origin is chosen so that this is zero, but not always.
  • \(v_{0y} = v_y(0)\text{:}\) This is the initial velocity in the vertical direction. If an object is dropped from rest or from a horizontally-moving vehicle, then this value will be zero.
  • \(y_0 = y(0)\text{:}\) This is the initial vertical position of the object. Most of the time, we take \(y = 0\) to be ground level.
We also have equations that give us the velocity in each direction.
\begin{align*} v_x(t) \amp = v_{0x} \\ v_y(t) \amp = -gt + v_{0y} \end{align*}
Notice that this means that the horizontal velocity is a constant. This is a reflection of an idea known as "Newton’s First Law of Motion", which we will explore in a later section.
The formulas above are completely general, but there is a special case that is worth independent investigation. This leads us to some additional formulas that are handy to have. However, it’s important to remmeber that this is a special case, and the formulas only apply when the assumptions have been met. Failing to remember these conditions leads to errors.

Dropped Objects.

We are often interested in the time it takes for a dropped object to hit the ground. Typically, this is done with the object at rest (not moving), which would mean that \(v_{0x} = 0\) and \(v_{0y} = 0\text{.}\) We will also assume that the initial position is \(x_0 = 0\) and \(y_0 = h\text{.}\) Under these assumptions, we have the following equations of motion:
\begin{align*} x(t) \amp = 0\\ y(t) \amp = -\frac{gt^2}{2} + h \end{align*}
Ultimately, we are interested in when the object hits the ground, which is \(y(t) = 0\text{.}\) We can plug this in and solve for \(t\text{:}\)
\begin{align*} 0 \amp = -\frac{gt^2}{2} + h\\ \frac{gt^2}{2} \amp = h\\ t^2 \amp = \frac{2h}{g} \\ t \amp = \sqrt{ \frac{2h}{g} } \end{align*}
We can also use this equation to determine the gravitational acceleration by dropping objects from various heights and measuring the time it takes for the object to hit the ground. We do this by solving for \(g\) instead of \(t\text{.}\)
\begin{align*} 0 \amp = -\frac{gt^2}{2} + h\\ \frac{gt^2}{2} \amp = h\\ g \amp = \frac{2h}{t^2} \end{align*}
In some cases, we don’t assume that \(v_{0x} = 0\text{,}\) such as when dropping an object from a moving vehicle (a car or a plane) or when throwing or launching an object horizontally. But this change does not affect the dynamics in the \(y\) coordinate.

A Launched Object from the Origin.

A common situation to consider is when an object is launched from the ground. We will assume that you know the initial horizontal and vertical velocities, but it’s possible with a little bit of trigonometry to be able to determine these from the total speed and the angle of launch. We also typically take \(x_0 = 0\) and \(y_0 = 0\text{,}\) which means that we are launching from ground level and we’re treating our launch position as the origin. With these substitutions, our equations become
\begin{align*} x(t) \amp = v_{0x} t\\ y(t) \amp = -\frac{gt^2}{2} + v_{0y}t \end{align*}
Using a bit of algebra, we can combine these equations together to determine the range of flight, which is how far the objecet travels before it hits the ground. To do this, we set \(y(t) = 0\) and solve for \(t\text{.}\) We get \(t = 0\) and \(t = \frac{2v_{0y}}{g}\text{.}\) The \(t = 0\) case is just our launch position, so we will focus on the other solution. This tells us that the object will be in the air until \(t_{\text{max}} = \frac{2v_{0y}}{g}\text{.}\) We can also take this value and plug it into the formula for \(x(t)\) to find out how far it traveled, giving \(x_{\text{max}} = v_{0x} \frac{2v_{0y}}{g} = \frac{2v_{0x}v_{0y}}{g}\text{.}\)
We can also use the time of flight to figure out how high the object the object went. The symmetrical shape of the parabola tells us that it will happen at the midpoint of flight, or at time \(t = \frac{1}{2} \cdot \frac{2v_{0y}}{g} = \frac{v_{0y}}{g}\text{.}\) If we plug this into the formula for the \(y\) position of the object, we get \(y_{\text{max}} = \frac{v_{0y}^2}{2g}\text{.}\)
Incidentally, we can get the same result from the vertical velocity equation. At the instant the object reaches maximum height, its vertical velocity is switching from positive to negative, which means that the vertical velocity is zero. If we set the vertical velocity equation equal to zero, we get \(v_y(t) = -gt + v_{0y} = 0 \text{,}\) and if we solve for \(t\text{,}\) we get \(t = \frac{v_{0y}}{g}\text{,}\) matching the value we derived above.
To summarize, when \(x_0 = 0\text{,}\) \(y_0 = 0\text{,}\) we have the following results:
  • Time of flight: \(t_{\text{max}} = \frac{2v_{0y}}{g}\)
  • Range of flight: \(x_{\text{max}} = \frac{2v_{0x}v_{0y}}{g}\)
  • Maximum height of flight: \(y_{\text{max}} = \frac{v_{0y}^2}{2g}\) at time \(\frac{1}{2} t_{\text{max}} = \frac{v_{0y}}{g}\)
It’s worth noting that in some cases, you will know the total initial speed \(v_0\) of the object and the launch angle, and you will need to derive the initial horizontal and vertical speeds. This is a simple application of trigonometry. If we draw a right triangle with hypotenuse length \(v_0\) and an angle \(\theta\) relative to the horizontal, then our trigonometry relationships tell the horizontal length is \(v_0 \cos(\theta)\) and the vertical length is \(v_{0y} = v_0 \sin(\theta)\text{.}\) In other words, we have \(v_{0x} = v_0 \cos(\theta)\)) and \(v_{0y} = v_0 \sin(\theta)\text{.}\) You will often see the projectile motion formulas presented in this way.

Practical Realities.

These formulas assume that we’ve removed the effect of air resistance. But it’s worth thinking about what would happen if we looked at more realistic physics that takes air resistance into account.
The first observation to make is that objects are not likely to go as high or as far. The reason is that the air resistance will tend to slow down the motion of the projectile. You can see this in practice by throwing a crumpled piece of paper. Unless the paper is wadded up very tightly, it will not go as far as you would expect when you throw it.
Another feature of adding air resistance is that there is a terminal velocity for falling objects. In the acceleration model we developed, there’s no limit to how fast something falls. However, the reality is that moving through the air will cause objects to reach a maximum speed when falling. Conceptually, you can think of this being the result of the object having to push air molecules out of the way, which takes away from the speed that it can attain.

Activity 5.1. Measuring \(g\).

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Instructions.

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