Welcome to Projectiles: Mastering 2D Motion!
Hello! Projectiles might sound complicated, but it's just a fantastic application of the straight-line motion you learned in M1, now applied in two directions at once.
In this chapter, you will learn how to analyze the path of any object launched into the air—whether it's a football, a cannonball, or just a stone thrown from a cliff. It's the physics behind everything from throwing a basketball to sending a rocket!
What is a Projectile?
A Projectile is any object that is given an initial velocity and then moves only under the influence of gravity.
Section 1: The Essential Assumptions (The Rules of the Game)
Before we start calculating, we must define the simplified environment we are working in. These assumptions are critical and often form part of exam questions.
1.1 Modelling Assumptions
When dealing with standard projectile motion problems in this course, we always make the following assumptions:
- The object is a particle: This means we ignore its size, shape, and rotational effects.
- Air Resistance is ignored: We assume the air has no effect on the particle's motion (a major simplification!).
- Gravity is constant: The acceleration due to gravity, \(g\), is constant throughout the motion (usually taken as \(g = 9.8 \text{ ms}^{-2}\)).
- Earth is flat: We assume motion occurs over a small enough distance that the Earth's curvature can be ignored, meaning \(g\) always acts vertically downwards.
Quick Review Box: Why Assumptions Matter
These assumptions allow us to treat the motion as having constant acceleration (\(g\) vertically) and zero acceleration (horizontally), which means we can use our reliable SUVAT equations!
Section 2: Separating the Motion (Horizontal vs. Vertical)
This is the single most important technique in solving projectile problems. We analyze the motion in the horizontal (x) direction and the vertical (y) direction completely independently.
2.1 Resolving Initial Velocity
Suppose a particle is launched from the origin with an initial speed \(V\) at an angle \(\alpha\) to the horizontal. We must first find the components of the initial velocity:
- Horizontal Component (\(u_x\)): \(u_x = V \cos \alpha\)
- Vertical Component (\(u_y\)): \(u_y = V \sin \alpha\)
2.2 The SUVAT Split
We now apply the SUVAT rules to each direction separately. Remember, time \(t\) is the only quantity common to both directions.
Horizontal Motion (X-Direction)
Since we ignore air resistance, there are no horizontal forces acting on the particle (Newton's First Law!).
- Acceleration: \(a_x = 0\)
- Initial Velocity: \(u_x = V \cos \alpha\)
- Displacement: \(x\)
Because \(a_x = 0\), the horizontal velocity is constant. The only SUVAT equation we need is:
$$x = u_x t \implies x = (V \cos \alpha) t$$
Vertical Motion (Y-Direction)
This motion is under the influence of gravity alone. We usually define the positive direction as upwards.
- Acceleration: \(a_y = -g\) (It's negative because gravity pulls down, opposite to the positive direction).
- Initial Velocity: \(u_y = V \sin \alpha\)
- Displacement: \(y\)
Using the SUVAT equation \(s = ut + \frac{1}{2} a t^2\):
$$y = u_y t + \frac{1}{2} a_y t^2 \implies y = (V \sin \alpha) t - \frac{1}{2} g t^2$$
Key Takeaway: Projectile motion is simplified into two standard equations:
$$x = (V \cos \alpha) t$$
$$y = (V \sin \alpha) t - \frac{1}{2} g t^2$$
Section 3: Key Calculations (Time, Height, and Range)
You must be able to calculate the following characteristics of the motion. The syllabus requires that you can derive the formulae for these quantities.
3.1 Time of Flight (\(T\))
The Time of Flight is the total time the projectile spends in the air, usually before it returns to its starting height (i.e., when \(y=0\)).
Step-by-Step Calculation:
- Set the vertical displacement \(y\) to zero:
$$(V \sin \alpha) t - \frac{1}{2} g t^2 = 0$$ - Factor out \(t\):
$$t \left( V \sin \alpha - \frac{1}{2} g t \right) = 0$$ - This gives two solutions: \(t=0\) (the start) or:
$$V \sin \alpha = \frac{1}{2} g t$$ - Solve for \(t\):
$$T = \frac{2 V \sin \alpha}{g}$$
Did you know? If you throw a ball straight up with initial speed \(u\), the time taken is \(2u/g\). Notice how the Time of Flight is simply \(2 u_y / g\)? This shows how vertical motion dictates the time!
3.2 Maximum Height (\(H_{max}\))
The projectile reaches its maximum height when its vertical velocity (\(v_y\)) is momentarily zero.
Step-by-Step Calculation:
- Find the time \(t_{peak}\) at which \(v_y = 0\). Use \(v = u + at\):
$$0 = (V \sin \alpha) + (-g) t_{peak} \implies t_{peak} = \frac{V \sin \alpha}{g}$$
Notice this is exactly half the total Time of Flight \(T\). - Substitute \(t_{peak}\) into the vertical displacement equation, \(y = (V \sin \alpha) t - \frac{1}{2} g t^2\):
$$H_{max} = (V \sin \alpha) \left( \frac{V \sin \alpha}{g} \right) - \frac{1}{2} g \left( \frac{V \sin \alpha}{g} \right)^2$$ - Simplify:
$$H_{max} = \frac{V^2 \sin^2 \alpha}{g} - \frac{1}{2} g \frac{V^2 \sin^2 \alpha}{g^2}$$ $$H_{max} = \frac{V^2 \sin^2 \alpha}{g} - \frac{V^2 \sin^2 \alpha}{2g}$$ $$H_{max} = \frac{V^2 \sin^2 \alpha}{2g}$$
3.3 Horizontal Range (\(R\))
The Range is the total horizontal distance covered when the projectile returns to its starting height.
Step-by-Step Calculation:
- Substitute the total Time of Flight \(T\) into the horizontal distance equation, \(x = (V \cos \alpha) t\):
$$R = (V \cos \alpha) \left( \frac{2 V \sin \alpha}{g} \right)$$ - Simplify using the double angle identity \(2 \sin \alpha \cos \alpha = \sin(2\alpha)\):
$$R = \frac{V^2 (2 \sin \alpha \cos \alpha)}{g}$$ $$R = \frac{V^2 \sin(2\alpha)}{g}$$
Common Mistake Alert!
Do NOT memorize the final formulae for \(T\), \(H_{max}\), or \(R\). You must be able to derive them step-by-step using the standard \(x\) and \(y\) equations. If you simply quote the final formula, you might lose method marks!
Section 4: The Trajectory Equation (The Path Itself)
Sometimes we want to find the equation that describes the path of the particle (its trajectory) without time \(t\) being involved. This equation relates \(y\) directly to \(x\).
4.1 Deriving the Trajectory Equation
We use substitution to eliminate \(t\) between the two primary equations:
1. Horizontal: \(x = (V \cos \alpha) t\)
2. Vertical: \(y = (V \sin \alpha) t - \frac{1}{2} g t^2\)
Step-by-Step Derivation:
- From (1), isolate \(t\):
$$t = \frac{x}{V \cos \alpha}$$ - Substitute this expression for \(t\) into equation (2):
$$y = (V \sin \alpha) \left( \frac{x}{V \cos \alpha} \right) - \frac{1}{2} g \left( \frac{x}{V \cos \alpha} \right)^2$$ - Simplify the first term using \(\frac{\sin \alpha}{\cos \alpha} = \tan \alpha\):
$$y = x \tan \alpha - \frac{g x^2}{2 V^2 \cos^2 \alpha}$$ - Use the trigonometric identity \(\frac{1}{\cos^2 \alpha} = \sec^2 \alpha\) to write the equation in its standard form:
$$y = x \tan \alpha - \frac{g x^2 \sec^2 \alpha}{2 V^2}$$
What does the Trajectory Equation tell us?
The equation \(y = (x \tan \alpha) - (\text{constant}) x^2\) shows that the trajectory of a projectile is always a parabola (a quadratic shape).
You might need to use this derived equation if the problem involves finding:
- Whether the particle hits a target at position \((x, y)\).
- The angle \(\alpha\) or initial velocity \(V\) required to pass through a specific point.
Analogy: Imagine throwing a basketball. The trajectory equation is like a detailed map of the path the ball takes, while the Time of Flight and Range are just specific points on that map.
Section 5: Advanced Projectile Problems
5.1 Finding Velocity at any Time
To find the velocity \(\mathbf{v}\) at any time \(t\), you need to calculate both the horizontal and vertical velocity components.
- Horizontal Velocity (\(v_x\)): Since \(a_x = 0\), this is always constant:
$$v_x = V \cos \alpha$$ - Vertical Velocity (\(v_y\)): Use \(v = u + at\):
$$v_y = V \sin \alpha - g t$$
The resultant speed \(|\mathbf{v}|\) is found using Pythagoras:
$$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$$
The direction \(\beta\) (angle with the horizontal) is found using trigonometry:
$$\tan \beta = \frac{|v_y|}{|v_x|}$$
5.2 Solving Non-Standard Problems
Some problems involve starting from a height (e.g., throwing a ball off a cliff).
For a struggling student, always go back to the basic definitions:
- Resolve V: Find \(u_x\) and \(u_y\).
- Define the Origin: Usually the point of launch \((0, 0)\).
- Set up Y: If the particle lands below the origin, its final vertical displacement \(y\) will be negative. This is the key difference!
- Use Time: Solve the vertical equation for time \(t\), and then substitute that time into the horizontal equation \(x = (V \cos \alpha) t\).
Memory Aid: SUVAT Check
When you are solving a tricky problem, mentally checklist your SUVAT inputs for the vertical direction:
U (Initial Velocity): \(V \sin \alpha\)
V (Final Velocity): Use only if known (e.g., \(v_y=0\) at peak).
A (Acceleration): \(-g\) (if upward is positive)
T (Time): The link!
Key Takeaway: All projectile problems, no matter how complex, boil down to correctly applying constant acceleration principles separately to the horizontal and vertical directions. Time is the bridge!