Welcome to the World of Projectiles!

In this chapter, we are going to explore the math behind objects flying through the air—from a kicked football to a launched satellite (well, almost!). In Further Mechanics, we take the basics you learned in your standard Mechanics course and look at the path (the "trajectory") in much more detail. Don't worry if you find the formulas a bit long; we’ll break them down step-by-step.

Prerequisite Check: Before we start, remember your basic SUVAT equations! We will be using them constantly. Specifically:
\( v = u + at \)
\( s = ut + \frac{1}{2}at^2 \)
\( v^2 = u^2 + 2as \)

1. Modeling the Motion

To make the math manageable, we use a simplified model. We treat the object as a particle. This means we ignore a few things that would happen in real life:

  • No air resistance: We imagine the object is moving in a vacuum.
  • No rotation: We don't care if the ball is spinning.
  • Constant gravity: We assume gravity \( (g) \) acts vertically downwards and doesn't change.

Did you know? In professional sports like golf or baseball, air resistance and "spin" (the Magnus effect) actually change the path significantly. However, for your 9231 exam, we stick to the idealized version!

Key Takeaway:

In our model, the horizontal acceleration is always zero and the vertical acceleration is always \( -g \) (taking upwards as positive).

2. Breaking it Down: Horizontal and Vertical

The secret to solving any projectile problem is to treat the horizontal motion and vertical motion as completely independent. Imagine two separate movies playing at the same time: one of a ball moving sideways at a constant speed, and one of a ball being thrown straight up and falling down.

Initial Velocity Components

If an object is launched with an initial speed \( u \) at an angle \( \theta \) to the horizontal:

  • Horizontal component (\( u_x \)): \( u \cos \theta \)
  • Vertical component (\( u_y \)): \( u \sin \theta \)

Horizontal Motion (No Acceleration)

Since there is no force pushing or pulling the object sideways:

Velocity: \( v_x = u \cos \theta \) (stays the same the whole time!)
Displacement: \( x = (u \cos \theta)t \)

Vertical Motion (Gravity Acts Here)

Gravity pulls the object down at \( g \approx 9.81 \, \text{m/s}^2 \):

Velocity: \( v_y = u \sin \theta - gt \)
Displacement: \( y = (u \sin \theta)t - \frac{1}{2}gt^2 \)

Quick Review: Horizontal = Constant Speed. Vertical = Constant Acceleration.

3. Important Milestones in the Flight

Most exam questions ask for one of these three "landmarks":

A. Greatest Height (\( H \))

At the very top of the path, the projectile stops moving up for a split second before it starts moving down. This means the vertical velocity \( v_y = 0 \).

Using \( v^2 = u^2 + 2as \):
\( 0 = (u \sin \theta)^2 - 2gH \)
Greatest Height: \( H = \frac{u^2 \sin^2 \theta}{2g} \)

B. Time of Flight (\( T \))

This is how long the object stays in the air. If it lands at the same horizontal level it started from, the vertical displacement \( y = 0 \).

Using \( s = ut + \frac{1}{2}at^2 \):
\( 0 = (u \sin \theta)T - \frac{1}{2}gT^2 \)
Time of Flight: \( T = \frac{2u \sin \theta}{g} \)

C. Horizontal Range (\( R \))

This is the total horizontal distance traveled. We use the Time of Flight (\( T \)) in the horizontal distance equation.

\( R = (u \cos \theta) \times T = (u \cos \theta) \times \frac{2u \sin \theta}{g} \)
Using the trig identity \( 2 \sin \theta \cos \theta = \sin 2\theta \):
Range: \( R = \frac{u^2 \sin 2\theta}{g} \)

Common Mistake to Avoid: These specific formulas for \( H, T, \) and \( R \) only work if the object lands on the same horizontal plane it started from! If it's kicked off a cliff, you must go back to the basic equations for \( x \) and \( y \).

4. The Cartesian Equation of the Trajectory

Sometimes, we don't care about the time (\( t \)). We just want to know the shape of the path (the \( y \) position for any given \( x \) position). This is the Cartesian Equation.

Step-by-Step Derivation:
  1. Start with the horizontal equation: \( x = (u \cos \theta)t \).
  2. Rearrange to find \( t \): \( t = \frac{x}{u \cos \theta} \).
  3. Substitute this \( t \) into the vertical equation: \( y = (u \sin \theta)t - \frac{1}{2}gt^2 \).
  4. \( y = (u \sin \theta)(\frac{x}{u \cos \theta}) - \frac{g}{2}(\frac{x}{u \cos \theta})^2 \).

After simplifying (remembering \( \frac{\sin \theta}{\cos \theta} = \tan \theta \) and \( \frac{1}{\cos^2 \theta} = \sec^2 \theta \)):

The Equation: \( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \)

Or, using \( \sec^2 \theta = 1 + \tan^2 \theta \):
\( y = x \tan \theta - \frac{gx^2}{2u^2}(1 + \tan^2 \theta) \)

Memory Trick: Notice that the equation is in the form \( y = ax - bx^2 \). This is the equation of a parabola that opens downwards!

5. Problem Solving Tips

When you see a tricky projectile question, follow this checklist:

  • Draw a diagram: Mark the initial speed, the angle, and the positive directions.
  • List what you know: Is it the height \( y \)? Is it the distance \( x \)? Is it the angle \( \theta \)?
  • Choose your direction: If you are looking for how far it traveled, look at Horizontal. If you are looking for how long it was in the air, look at Vertical.
  • Link them with time: Time (\( t \)) is the "bridge" that connects horizontal and vertical motion.

Don't worry if this seems tricky at first! The key is practice. Once you realize that the horizontal speed never changes, half the battle is won.

Key Takeaway Summary:

1. Horizontal: \( a_x = 0 \), so \( v_x \) is constant.
2. Vertical: \( a_y = -g \), use SUVAT.
3. Trajectory: The path is a parabola defined by \( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \).