Mathematics - Further (9231) | Motion of a projectile

Welcome to the exciting world of Motion of a Projectile! This chapter takes your prior knowledge of kinematics (SUVAT) and extends it into two dimensions (horizontal and vertical motion simultaneously). Why is this important? Because anything you throw, kick, or fire follows these principles—from a cricket ball to a space shuttle launch.

The concepts here are fundamental to Further Mechanics (Paper 3). If you can master splitting vectors and applying the correct acceleration to each dimension, you’ve cracked the hardest part!

1. The Ideal Projectile Model: Assumptions are Key

To make the mathematics manageable, we use an ideal model. In Further Mechanics, understanding the assumptions of the model is just as important as solving the equations.

1.1 What is a Projectile?

A projectile is any object moving freely under the influence of gravity alone.

1.2 Key Assumptions and Limitations of the Model

We treat the projectile as a particle moving with constant acceleration.

• Assumption 1: Air Resistance is Ignored. In real life, drag slows things down, but in this ideal model, there is zero resistance.
• Assumption 2: Gravity is Constant. The acceleration due to gravity, \(g\), is assumed to be constant in both magnitude and direction throughout the flight. We typically use \(g \approx 9.8 \text{ m/s}^2\) (or sometimes \(10 \text{ m/s}^2\)).
• Assumption 3: Earth's Curvature is Ignored. For typical projectile ranges, we assume the ground is a flat, horizontal plane.

If a question asks for the limitations of the model, refer directly to these assumptions!

Quick Review: The motion is modelled as two separate, independent types of motion happening at the same time: one horizontal, one vertical.

2. Deconstructing the Motion (Horizontal and Vertical)

The core skill in projectile motion is resolving the initial velocity into its components and applying the appropriate SUVAT equations for each dimension.

2.1 Initial Velocity Components

Let the initial speed be \(V\) and the angle of projection above the horizontal be \(\theta\).

• Initial Horizontal Velocity: \(V_x = V \cos \theta\)
• Initial Vertical Velocity: \(V_y = V \sin \theta\)

2.2 Horizontal Motion (X-Direction)

The acceleration in the horizontal direction is zero (\(a_x = 0\)) because air resistance is ignored.

This means the horizontal motion is constant velocity.

If \(x\) is the horizontal displacement at time \(t\), then:

\(x = (\text{Constant Velocity}) \times (\text{Time})\)
\[x = (V \cos \theta) t\]

Memory Aid: Horizontal motion is easy—it’s just speed = distance / time, as long as time \(t\) is the same time used for the vertical motion.

2.3 Vertical Motion (Y-Direction)

The acceleration in the vertical direction is constant (\(a_y = -g\)), assuming upward is positive.

We use the standard SUVAT equations, replacing \(a\) with \(-g\).

• Vertical Velocity (\(v_y\)) at time \(t\): \[v_y = V \sin \theta - gt\]
• Vertical Displacement (\(y\)) at time \(t\): \[y = (V \sin \theta) t - \frac{1}{2} g t^2\]
• Velocity Relationship: \[v_y^2 = (V \sin \theta)^2 - 2gy\]

Important Note: Always define your positive direction (usually upwards is positive, making \(a = -g\)). Be consistent!

3. Standard Projectile Results and Calculations

You must be able to calculate three standard metrics: the time to reach maximum height, the greatest height itself, and the total range on a horizontal plane.

3.1 Greatest Height Reached (\(H\))

The greatest height occurs when the vertical velocity (\(v_y\)) is zero.

Step 1: Find the time to max height (\(t_H\)).
Using \(v_y = V \sin \theta - gt\) and setting \(v_y = 0\):

\[0 = V \sin \theta - g t_H\] \[t_H = \frac{V \sin \theta}{g}\]

Step 2: Find the maximum height (\(H\)).
Substitute \(t_H\) into the displacement equation \(y = (V \sin \theta) t - \frac{1}{2} g t^2\), or use the equation without \(t\):

Using \(v_y^2 = (V \sin \theta)^2 - 2gy\) and setting \(v_y = 0\) and \(y=H\):

\[0 = V^2 \sin^2 \theta - 2gH\] \[H = \frac{V^2 \sin^2 \theta}{2g}\]

3.2 Time of Flight (\(T\))

If the projectile lands on the same horizontal level from which it was projected, the total time of flight (\(T\)) is simply \(2 t_H\).

\[T = \frac{2V \sin \theta}{g}\]

Alternatively, the time of flight occurs when the vertical displacement \(y\) is zero (apart from \(t=0\)).

Using \(y = (V \sin \theta) t - \frac{1}{2} g t^2\) and setting \(y=0\):

\[0 = t \left( V \sin \theta - \frac{1}{2} g t \right)\]

This gives \(t=0\) (start) or \(T = \frac{2V \sin \theta}{g}\) (end).

3.3 Range (\(R\))

The Range (\(R\)) is the total horizontal distance travelled when the time elapsed is \(T\). We use the horizontal equation \(x = (V \cos \theta) t\):

\[R = (V \cos \theta) T = (V \cos \theta) \left( \frac{2V \sin \theta}{g} \right)\] \[R = \frac{V^2 (2 \sin \theta \cos \theta)}{g}\]

Using the double angle identity (\(2 \sin \theta \cos \theta = \sin 2\theta\)):

\[R = \frac{V^2 \sin 2\theta}{g}\]

Did you know? The maximum range for a fixed initial speed \(V\) occurs when \(\sin 2\theta = 1\), which means \(2\theta = 90^\circ\), or \(\theta = 45^\circ\). To throw something furthest, you aim at 45 degrees!

3.4 Velocity at Time \(t\) or Position \((x, y)\)

To find the velocity at any point, calculate both components:

\(v_x = V \cos \theta\) (constant)
\(v_y = V \sin \theta - gt\)

The magnitude of the velocity (speed) is:

\[|v| = \sqrt{v_x^2 + v_y^2}\]

The direction of the velocity (angle \(\phi\) with the horizontal) is:

\[\tan \phi = \frac{v_y}{v_x}\]

Common Mistake to Avoid: Remember that when the projectile is descending, \(v_y\) will be negative, meaning \(\phi\) will be negative (below the horizontal).

4. The Cartesian Equation of the Trajectory

The Cartesian equation of the trajectory describes the path of the projectile, relating the vertical position (\(y\)) directly to the horizontal position (\(x\)), without needing time (\(t\)).

4.1 Deriving the Trajectory Equation

This derivation is required and relies on eliminating the parameter \(t\).

Start with the two displacement equations:

1. Horizontal: \(x = (V \cos \theta) t\)
2. Vertical: \(y = (V \sin \theta) t - \frac{1}{2} g t^2\)

Step 1: Express \(t\) in terms of \(x\) from Equation 1.

\[t = \frac{x}{V \cos \theta}\]

Step 2: Substitute this expression for \(t\) into Equation 2.

\[y = (V \sin \theta) \left( \frac{x}{V \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{V \cos \theta} \right)^2\]

Step 3: Simplify the terms.

The first term simplifies using \(\frac{\sin \theta}{\cos \theta} = \tan \theta\):

\[y = x \tan \theta - \frac{g x^2}{2V^2 \cos^2 \theta}\]

4.2 Using the Trajectory Equation

This formula is provided on the MF19 list of formulae (using \(V\) and \(\theta\)): \[y = x \tan \theta - \frac{g x^2}{2V^2 \cos^2 \theta}\]

This equation is crucial when solving problems where the initial conditions (\(V\) or \(\theta\)) are unknown, but you know the coordinates \((x, y)\) of a specific point the projectile passes through (e.g., scoring a goal, clearing a wall).

Since the equation is quadratic in \(x\) (containing \(x\) and \(x^2\)), it confirms that the path of the projectile is a parabola.

Example Use: If a projectile passes through a point (10, 3) and you know \(V\) but not \(\theta\), substituting \(x=10\) and \(y=3\) gives you a single equation involving only \(\tan \theta\) and \(\cos^2 \theta\). You can use the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) (where \(\sec \theta = 1/\cos \theta\)) to solve for \(\theta\).

5. Step-by-Step Problem Solving Strategy

Projectile questions often look complicated, but following a consistent strategy helps break them down.

5.1 Checklist for Projectile Problems

1. Resolve: Split the initial velocity \(V\) into horizontal (\(V \cos \theta\)) and vertical (\(V \sin \theta\)) components.
2. List Accelerations: State \(a_x = 0\) and \(a_y = -g\) (if up is positive).
3. Identify Unknown: What is the question asking for? (Time? Height? Range? Angle?)
4. Select Dimension:
   • If the goal relates to distance and time, use \(x\) and \(y\) displacements.
   • If the goal relates to velocity, use \(v_x\) and \(v_y\).
   • If the goal relates to position without time, consider the Trajectory Equation.
5. The Time Link: If you need to find something horizontal (like range), you often need to solve the vertical motion first to find the total time of flight, \(T\).

Don't worry if this seems tricky at first; practice is everything. Always draw a diagram and label your components clearly!

5.2 When to Use the Trajectory Equation vs. SUVAT

• Use SUVAT (\(x\) and \(y\) equations with \(t\)): When time \(t\) is known or can be easily found (e.g., finding maximum height, finding velocity at 5 seconds).
• Use the Trajectory Equation: When time \(t\) is irrelevant or when you need to solve for an unknown initial condition (\(V\) or \(\theta\)) given a specific target coordinate \((x, y)\). This is often quicker when intersecting a fixed barrier or target.