Welcome to the World of Projectiles!
In your previous mechanics chapters, you looked at objects moving in straight lines—either left and right or up and down. In this chapter, we are going to level up! We will explore Motion under gravity in 2 dimensions. Think of a football being kicked, a stone being thrown, or a stunt car jumping over a gap. These objects move forward and up/down at the same time. This is also known as Projectile Motion.
Don't worry if this seems a bit more complex at first. The "secret" to mastering this topic is realizing that a 2D problem is just two 1D problems happening at the exact same time. Let's break it down!
1. The Rules of the Game: Modelling Assumptions
To make the math manageable at A Level, we use a few modelling assumptions. When you see a question about a "projectile," you can assume these things are true unless the question says otherwise:
The Core Assumptions:
• The object is a particle: We ignore its shape and any rotation (like the spin on a ball).
• No air resistance: We pretend the air doesn't push against the object.
• Constant gravity: We assume the acceleration due to gravity, \( g \), is always \( 9.8 \text{ m s}^{-2} \) and always acts vertically downwards.
• The Earth is flat: For the distances we calculate, we don't need to worry about the curvature of the Earth!
Quick Review: Why do we assume no air resistance? Because it makes the horizontal motion simple—without air pushing back, the horizontal speed never changes!
2. The Golden Rule: Independence of Motion
This is the most important concept in this chapter: Horizontal and Vertical motions are completely independent.
Imagine you drop a ball at the exact same moment you fire another ball horizontally. Did you know? They will both hit the ground at the same time! This is because gravity only pulls things down; it doesn't care how fast they are moving sideways.
A. Horizontal Motion (\( x \)-direction)
Because there is no air resistance and no horizontal force, there is no horizontal acceleration (\( a_x = 0 \)).
• Horizontal Velocity: This stays constant throughout the flight.
• Equation to use: \( \text{Distance} = \text{Speed} \times \text{Time} \) or \( x = (u \cos \theta)t \).
B. Vertical Motion (\( y \)-direction)
Gravity is always pulling the object down, so there is a constant vertical acceleration (\( a_y = -9.8 \text{ m s}^{-2} \), assuming upwards is positive).
• Vertical Velocity: This changes every second.
• Equation to use: Use your SUVAT equations!
Key Takeaway: Always split your page into two columns—one for Horizontal and one for Vertical. Never mix their numbers!
3. Setting Up the Vectors
Most problems start with an object launched at an initial velocity \( u \) at an angle \( \theta \) to the horizontal. We need to split this into components:
• Horizontal component (\( u_x \)): \( u \cos \theta \)
• Vertical component (\( u_y \)): \( u \sin \theta \)
Memory Aid: "Cos is across" (Horizontal). If you move across the angle, use Cosine. If you move away from the angle, use Sine.
In vector notation, we can write the initial velocity as:
\( \mathbf{u} = \begin{pmatrix} u \cos \theta \\ u \sin \theta \end{pmatrix} \)
And the acceleration as:
\( \mathbf{a} = \begin{pmatrix} 0 \\ -g \end{pmatrix} \)
4. Finding Position and Velocity at any Time \( t \)
Using the vector versions of SUVAT, we can find where the object is (its position vector \( \mathbf{r} \)) and how fast it’s going (its velocity vector \( \mathbf{v} \)) at any point in time.
Velocity at time \( t \):
\( \mathbf{v} = \mathbf{u} + \mathbf{a}t = \begin{pmatrix} u \cos \theta \\ u \sin \theta - gt \end{pmatrix} \)
Displacement at time \( t \):
\( \mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 = \begin{pmatrix} (u \cos \theta)t \\ (u \sin \theta)t - \frac{1}{2}gt^2 \end{pmatrix} \)
Common Mistake to Avoid: Don't forget the minus sign for \( g \)! If you've decided that "up" is the positive direction, gravity must be negative because it acts "down."
5. Key Projectile Milestones
In exams, you are often asked for three specific things:
I. Maximum Height (\( H \))
At the very top of the arc, the object stops moving up for a split second before it starts moving down. This means the vertical velocity is zero (\( v_y = 0 \)).
Step-by-step: Use \( v^2 = u^2 + 2as \) in the vertical column with \( v_y = 0 \).
II. Time of Flight (\( T \))
This is how long the object is in the air. If the object starts and ends at the same horizontal level, the vertical displacement is zero (\( s_y = 0 \)).
Step-by-step: Use \( s = ut + \frac{1}{2}at^2 \) in the vertical column with \( s_y = 0 \).
III. Horizontal Range (\( R \))
This is the total horizontal distance travelled. It is simply the constant horizontal speed multiplied by the total time of flight.
Step-by-step: \( R = (u \cos \theta) \times T \).
Key Takeaway: The Time (\( t \)) is the only variable that is the same for both horizontal and vertical motions. It's the "bridge" that connects your two columns of math!
6. The Equation of the Trajectory
Sometimes we want to know the path the projectile takes on a graph (the \( y \) position for any given \( x \) position) without caring about the time.
To do this, we eliminate \( t \) from our equations:
1. From the horizontal motion: \( t = \frac{x}{u \cos \theta} \)
2. Substitute this \( t \) into the vertical displacement equation: \( y = (u \sin \theta)t - \frac{1}{2}gt^2 \)
After some trig simplification (using \( \frac{\sin \theta}{\cos \theta} = \tan \theta \)), you get the Trajectory Equation:
\( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \)
Shape Alert! Because this equation has an \( x^2 \) term with a negative coefficient, the path is always an inverted parabola (an upside-down 'U' shape).
7. Problem Solving Strategy: A Quick Review
Feeling overwhelmed? Follow these steps for any projectile problem:
- Draw a diagram: Mark your launch point, the angle, and the initial speed.
- Pick your directions: Usually, let up be positive and right be positive.
- Resolve initial velocity: Find \( u_x = u \cos \theta \) and \( u_y = u \sin \theta \).
- Split your math: Create a "Horizontal" column (Speed, Distance, Time) and a "Vertical" column (SUVAT).
- Find Time (\( t \)): Use the column where you have the most information to find \( t \).
- Bridge across: Use that \( t \) in the other column to find your final answer.
One Last Encouragement: Projectile motion is just a puzzle. As long as you keep your horizontal and vertical numbers in their own columns, the pieces will always fit together!