Welcome to the World of Projectiles!
Ever wondered how a basketball player knows exactly how to arc a shot, or how a rescue package dropped from a plane reaches its target? In this chapter, we explore Projectiles—the study of objects moving through the air under the influence of gravity alone. This is one of the most exciting parts of the AQA Mechanics course because it combines your knowledge of vectors and kinematics to describe motion in two dimensions.
Don’t worry if 2D motion sounds intimidating! The "secret trick" to projectiles is that we look at horizontal and vertical movement completely separately. Let’s dive in.
1. The Modelling Assumptions
To make the maths manageable, we use a mathematical model (Theme OT3). In AQA Level Mathematics, we usually assume:
1. The object is a particle (its size and shape don't matter).
2. There is no air resistance.
3. The acceleration due to gravity (\(g\)) is constant and acts vertically downwards (\(g \approx 9.8 \, \text{ms}^{-2}\)).
4. The motion happens in a vertical plane.
Quick Tip: Because we ignore air resistance, the horizontal speed of a projectile never changes until it hits something!
2. Breaking Down the Motion
The most important rule in projectiles is: Horizontal and vertical motions are independent.
Horizontal Motion (\(x\))
Since there are no horizontal forces (no air resistance!), the horizontal acceleration is zero (\(a_x = 0\)).
Horizontal velocity (\(v_x\)) is constant.
Formula: \(x = u_x \times t\)
Vertical Motion (\(y\))
Gravity is the only force acting here, so the vertical acceleration is constant (\(a_y = -g\)).
We use the standard SUVAT equations for vertical motion.
Key Formula: \(y = u_y t - \frac{1}{2}gt^2\)
3. Starting the Journey: Initial Velocity
Usually, a projectile is launched at an initial speed \(U\) at an angle \(\theta\) to the horizontal. Before you do anything else, you must resolve this velocity into components:
Horizontal component: \(u_x = U \cos \theta\)
Vertical component: \(u_y = U \sin \theta\)
Analogy: Think of the horizontal component as how fast the shadow of the ball moves along the ground, and the vertical component as how fast the ball would go if you threw it straight up.
4. Using Vectors in Projectiles
The AQA syllabus emphasizes using vectors (\(\mathbf{i}\) and \(\mathbf{j}\) notation) to solve these problems (Section Q). This is often the cleanest way to work!
If \(\mathbf{i}\) is the unit vector in the horizontal direction and \(\mathbf{j}\) is the unit vector vertically upwards, we can write the acceleration vector as:
\(\mathbf{a} = 0\mathbf{i} - g\mathbf{j}\)
The position vector (\(\mathbf{r}\)) at any time \(t\) is given by:
\(\mathbf{r} = (U \cos \theta)t \mathbf{i} + \left( (U \sin \theta)t - \frac{1}{2}gt^2 \right) \mathbf{j}\)
Key Takeaway: The \(\mathbf{i}\) component tells you the horizontal distance (range), and the \(\mathbf{j}\) component tells you the height.
5. Important Moments in Flight
There are three common "events" you will be asked to calculate:
Maximum Height
At the very top of the arc, the projectile stops going up for a split second before it starts coming down.
Condition: Vertical velocity is zero (\(v_y = 0\)).
Time of Flight
This is how long the object is in the air. If it is launched from the ground and lands on the ground:
Condition: Vertical displacement is zero (\(y = 0\)).
The Range
This is the total horizontal distance travelled. To find this, first calculate the Time of Flight, then multiply it by the constant horizontal velocity (\(u_x\)).
6. Common Mistakes to Avoid
1. Mixing up Sine and Cosine: Remember, "Cos is across" (horizontal) for the angle with the ground.
2. Signs for Gravity: If you define upwards as positive, then \(a = -9.8\). If you forget the minus sign, your projectile will accelerate into space!
3. Combining Components: Never try to use the total initial speed \(U\) in a SUVAT equation for vertical motion. Only use the vertical component \(U \sin \theta\).
4. Calculator Units: Ensure your calculator is in Degrees mode unless the question specifies Radians (Section E).
7. Problem-Solving Step-by-Step
When you see a projectile question, follow these steps:
Step 1: Draw a quick sketch. Label the launch point as \((0,0)\).
Step 2: Resolve the initial velocity into \(U \cos \theta\) and \(U \sin \theta\).
Step 3: List your SUVAT variables for both horizontal and vertical directions.
Step 4: Identify what you need to find (e.g., "How far..." means horizontal distance; "How high..." means vertical distance).
Step 5: Choose the appropriate equation and solve!
Did you know? In real life, air resistance makes the path of a projectile "steeper" on the way down and significantly reduces the range. This is why our mathematical models are "simplified" (OT3).
Summary Table: The Projectile Split
Horizontal (\(\mathbf{i}\))
Acceleration: \(a = 0\)
Velocity: \(v = u \cos \theta\) (Constant)
Displacement: \(x = (u \cos \theta)t\)
Vertical (\(\mathbf{j}\))
Acceleration: \(a = -g\)
Velocity: \(v = u \sin \theta - gt\)
Displacement: \(y = (u \sin \theta)t - \frac{1}{2}gt^2\)
Final Encouragement: Projectiles might seem like they have a lot of moving parts, but it is really just two simple 1D problems happening at the same time. Master the split, and you’ll master the chapter!