Projectile Motion: Parametric Equations

Hello! Welcome to your next lesson on kinematics.

In our previous session, we established the cornerstone of multidimensional kinematics: the principle of independence for perpendicular components of motion. You rigorously proved that for a fixed Cartesian coordinate system, the kinematic equations for the x and y directions form two uncoupled sets of differential equations, linked only by the common parameter of time, $t$ .

Today, we will apply this principle to the most classic and crucial example of 2D motion: the motion of a projectile. Your learning outcome is to derive the parametric equations of motion (x(t), y(t)) for a projectile under constant gravity. We will approach this from first principles using vector calculus, a method that aligns with your preference for rigorous mathematical derivations.

1. Defining the Physical Model of Projectile Motion

Projectile motion describes the path of an object that is given an initial velocity and subsequently moves only under the influence of gravity. To build a mathematical model, we make a key idealization:

Assumption: We neglect air resistance and the rotation of the Earth.
Consequence: The only force acting on the projectile is its weight, which results in a constant downward acceleration.

We'll establish a standard 2D Cartesian coordinate system where the positive y-axis points vertically upwards and the positive x-axis is horizontal. In this frame, the acceleration due to gravity, $g$ , is always directed towards the center of the Earth (downwards).

Therefore, the acceleration vector $a$ of the projectile is constant and can be written as:
$a (t) = 0 i^- g j^$
where $g$ is the magnitude of the acceleration due to gravity (approximately $9.8 m/s^{2}$ ).

Notice that $a_{x} = 0$ and $a_{y} = - g$ . This is the starting point for our entire derivation.

2. Derivation via Vector Integration

Since you are comfortable with calculus, we will derive the equations of motion by integrating the acceleration vector. This is the most fundamental approach.

Step 1: From Acceleration to Velocity

We begin with the definition of acceleration:
$a (t) = \frac{d v}{d t}$
To find the velocity vector $v (t)$ , we integrate $a (t)$ with respect to time:
$v (t) = \int a (t) d t = \int (- g j^) d t$
Since $j^$ is a constant vector, the integration yields:
$v (t) = (- g t) j^+ C_{1}$
Here, $C_{1}$ is a constant vector of integration. To determine its value, we apply the initial conditions. Let the velocity at $t = 0$ be the initial velocity vector, $v_{0}$ .
$v (0) = v_{0} = (- g \cdot 0) j^+ C_{1} ⟹ C_{1} = v_{0}$
Thus, the velocity vector at any time $t$ is:
$v (t) = v_{0} - g t j^$

Step 2: From Velocity to Position

Next, we use the definition of velocity:
$v (t) = \frac{d r}{d t}$
To find the position vector $r (t)$ , we integrate $v (t)$ with respect to time:
$r (t) = \int v (t) d t = \int (v_{0} - g t j^) d t$
Since $v_{0}$ is a constant vector, this gives:
$r (t) = v_{0} t - \frac{1}{2} g t^{2} j^+ C_{2}$
$C_{2}$ is our second constant vector of integration. We find it by applying the initial position. Let the position at $t = 0$ be $r_{0}$ .
$r (0) = r_{0} = v_{0} (0) - \frac{1}{2} g (0)^{2} j^+ C_{2} ⟹ C_{2} = r_{0}$
Substituting this back, we arrive at the general vector equation for projectile motion:
$r (t) = r_{0} + v_{0} t - \frac{1}{2} g t^{2} j^$

This elegant vector equation contains all the information about the projectile's path. To extract the parametric equations, we simply break it down into its components.

3. The Parametric Equations of Motion

Let's express the initial conditions in component form. Assume the projectile is launched from position $r_{0} = x_{0} i^+ y_{0} j^$ with an initial velocity $v_{0}$ that has magnitude $v_{0}$ and is directed at an angle $θ$ above the horizontal.

Initial velocity components: $v_{0} = (v_{0} cos θ) i^+ (v_{0} sin θ) j^$

Now, we substitute these components into our general vector equation for $r (t)$ :
$r (t) = (x_{0} i^+ y_{0} j^) + ((v_{0} cos θ) i^+ (v_{0} sin θ) j^) t - \frac{1}{2} g t^{2} j^$
To get the parametric equations, we group the $i^$ and $j^$ terms. The position vector is $r (t) = x (t) i^+ y (t) j^$ .
$x (t) i^+ y (t) j^= (x_{0} + (v_{0} cos θ) t) i^+ (y_{0} + (v_{0} sin θ) t - \frac{1}{2} g t^{2}) j^$

By comparing the components on both sides, we obtain the parametric equations for projectile motion:

$x (t) = x_{0} + (v_{0} cos θ) t$
$y (t) = y_{0} + (v_{0} sin θ) t - \frac{1}{2} g t^{2}$

These two equations fully describe the projectile's position at any time $t$ .

The following document provides a concise, formal proof that is very similar to what we have just done.

Integration and projectile motion (Sect. 13.2)

The document 'Integration and projectile motion' presents a rigorous mathematical derivation of the projectile motion equations. Please focus on the section that mirrors our derivation.

Please read the section titled 'Equations of a projectile motion', starting from the theorem about the motion of a particle. The proof shown there starts from r''(t) = -gj and integrates twice, just as we did, to arrive at the final component equations. This will solidify your understanding of the formal derivation.

You can also watch this video for a dynamic walkthrough of the same integration process.

Projectile Motion: A Vector Calculus Problem

The video 'Projectile Motion: A Vector Calculus Problem' provides a clear, step-by-step visual demonstration of integrating the acceleration vector to find the velocity and position vectors.

Watch the segment from 01:31 to 04:03. This part covers the integration of the acceleration vector to get velocity, applying initial conditions, and then integrating the velocity vector to obtain the position vector and its corresponding parametric equations.

4. Connection to the Principle of Independence

Notice that our final parametric equations are exactly what we would get by applying the Principle of Independence of Motion directly, using the 1D constant-acceleration kinematic equations you derived in Module 2.

X-Motion: The acceleration is $a_{x} = 0$ . This implies uniform motion.
- The 1D equation is $x (t) = x_{0} + v_{0 x} t$ .
- Substituting $v_{0 x} = v_{0} cos θ$ gives: $x (t) = x_{0} + (v_{0} cos θ) t$ .
Y-Motion: The acceleration is $a_{y} = - g$ . This is uniformly accelerated motion.
- The 1D equation is $y (t) = y_{0} + v_{0 y} t + \frac{1}{2} a_{y} t^{2}$ .
- Substituting $v_{0 y} = v_{0} sin θ$ and $a_{y} = - g$ gives: $y (t) = y_{0} + (v_{0} sin θ) t - \frac{1}{2} g t^{2}$ .

Both the direct application of 1D equations and the more fundamental vector integration yield the same result. This demonstrates the consistency of the physics framework we are building.

The image below provides an excellent summary of the vector and parametric equations for a projectile launched from the origin.

Vector and Parametric Equations for Ideal Projectile Motion — This diagram summarizes the derivation of the parametric equations. It shows the initial conditions, the general vector equation for position $\vec{r}(t)$, and the final component equations $x(t)$ and $y(t)$ for a projectile launched from the origin ($x_0=0, y_0=0$).

Test your understanding!

A cannonball is fired from a cliff 50 meters high ( $y_{0} = 50$ ). Its initial position is $x_{0} = 0$ . The initial speed is $v_{0} = 100$ m/s at an angle $θ = 3 0^{\circ}$ above the horizontal. Using $g = 10 m/s^{2}$ , write down the specific parametric equations $x (t)$ and $y (t)$ for this cannonball's motion.

Show answer

First, we calculate the initial velocity components:

$v_{0 x} = v_{0} cos θ = 100 cos (3 0^{\circ}) = 100 \cdot \frac{3}{2} = 503$ m/s.
$v_{0 y} = v_{0} sin θ = 100 sin (3 0^{\circ}) = 100 \cdot \frac{1}{2} = 50$ m/s.

Now we plug these values, along with $x_{0} = 0$ , $y_{0} = 50$ , and $g = 10$ , into our general parametric equations:

$x (t) = x_{0} + v_{0 x} t ⟹ x (t) = (503) t$
$y (t) = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} ⟹ y (t) = 50 + 50 t - 5 t^{2}$

These two equations define the cannonball's position at any time $t > 0$ (until it hits the ground).

Conclusion

In this lesson, you have derived the fundamental equations governing projectile motion from first principles.

Key Takeaways:

Projectile motion is modeled by assuming a constant downward gravitational acceleration $a = - g j^$ .
By integrating the acceleration vector twice and applying initial conditions, we derived the general position vector: $r (t) = r_{0} + v_{0} t - \frac{1}{2} g t^{2} j^$ .
Decomposing this vector equation gives the parametric equations of motion:
- $x (t) = x_{0} + (v_{0} cos θ) t$ (Uniform motion)
- $y (t) = y_{0} + (v_{0} sin θ) t - \frac{1}{2} g t^{2}$ (Uniformly accelerated motion)
This derivation is a direct and powerful application of vector calculus and the principle of independent components.

Preview of the Next Lesson:
The parametric equations $x (t)$ and $y (t)$ tell us where the projectile is at a given time. But what is the shape of its path? In the next lesson, we will eliminate the parameter $t$ from these two equations to derive the trajectory equation, $y (x)$ . We will then rigorously prove that this equation describes a parabola.

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