# Projectile Motion

> Why a thrown ball follows a curve, and how to split its motion into a steady sideways part and a falling part.

From HSC Physics on WhizzNotes, free study notes for NSW students: https://whizznotes.com/physics/projectile-motion/. Updated 2 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Two balls, one table
Mia rolls a marble off the edge of her desk. At the exact same moment, her brother Leo simply lets go of a second marble from the same height.

"Mine will land first," says Leo. "It has a shorter path."

They try it ten times and listen. Every time there is a single *click*. Both marbles hit the floor together.

Mia's marble moves sideways while it falls, but the sideways motion does not change how fast it falls. Gravity pulls both marbles down in exactly the same way. This one idea is the key to every projectile question.

**In short:** A projectile is an object that moves under gravity alone after it is launched. Its horizontal velocity stays constant while its vertical velocity changes at 9.8 m/s² downwards, so the two directions are solved separately and linked by time.

## What is a projectile?

A **projectile** is any object that is thrown, kicked or launched, and then moves under **gravity alone**. Examples: a netball shot, a long jumper in the air, water from a hose, a stone flicked off a cliff.

In this chapter we **ignore air resistance**. Then the only force on the projectile is its **weight**, which points straight down. So:

- The acceleration is $g = 9.8$ m/s² **downwards**, the whole time.
- There is **no** horizontal force, so there is **no** horizontal acceleration.

The path is a curve called a **parabola**.

## Split the motion in two

The trick is to treat the projectile as two separate motions that happen at the same time.

| | Horizontal ($x$) | Vertical ($y$) |
|---|---|---|
| Force | none | weight, down |
| Acceleration | $0$ | $9.8$ m/s² down |
| Velocity | **constant** | changes by 9.8 m/s every second |
| Equation | $s_x = u_x t$ | $s_y = u_y t + \frac{1}{2} a_y t^2$ |

The only thing the two parts share is **time**. The time the ball spends in the air is the same for both.

### Key idea: Projectile motion in one line
Sideways: steady speed. Up and down: falling with $g$. Same time for both.

## Breaking the launch velocity into parts

A ball launched at speed $u$ at angle $\theta$ above the horizontal has two **components**:

$$u_x = u\cos\theta \qquad u_y = u\sin\theta$$

### Worked example: A kicked football
A football is kicked at 20 m/s at 30° above the ground.

- $u_x = 20\cos 30° = 17.3$ m/s (this stays the same for the whole flight)
- $u_y = 20\sin 30° = 10$ m/s (upwards at the start)

## The useful equations

Take **up as positive**, so $a_y = -9.8$ m/s².

- Vertical velocity at time $t$: $v_y = u_y + a_y t$
- Vertical position: $s_y = u_y t + \frac{1}{2}a_y t^2$
- Without time: $v_y^2 = u_y^2 + 2a_y s_y$
- Horizontal position: $s_x = u_x t$

## Key moments in a flight

**At the top (maximum height):**

- Vertical velocity is **zero**.
- Horizontal velocity is still $u_x$, so the ball is **not stopped**. Its speed is $u_x$.
- Acceleration is still 9.8 m/s² down.

**Landing at the same height it started:**

- The flight is **symmetrical**. Time up = time down.
- It lands with the same speed it was launched with, at the same angle below the horizontal.

### Worked example: Time, height and range
A ball is launched at 20 m/s at 30° and lands at the same height. Use $g = 9.8$ m/s².

1. Components: $u_x = 17.3$ m/s, $u_y = 10$ m/s.
2. Time to the top: $0 = 10 - 9.8t$, so $t = 1.02$ s.
3. Total time of flight: $2 \times 1.02 = 2.04$ s.
4. Maximum height: $0 = 10^2 - 2(9.8)h$, so $h = 5.1$ m.
5. Range: $s_x = 17.3 \times 2.04 = 35.3$ m.

## Launched horizontally

When something is thrown straight out sideways, $u_y = 0$. The fall time depends **only on the height**:

$$h = \tfrac{1}{2} g t^2$$

Throwing it faster makes it land **further away**, but it does **not** change the time.

### Worked example: Off a cliff
A stone is thrown horizontally at 12 m/s from a 45 m cliff.

- Fall time: $45 = \frac{1}{2}(9.8)t^2$, so $t = 3.03$ s.
- Range: $12 \times 3.03 = 36.4$ m.

## Changing the launch angle

For the same launch speed on level ground:

- **45°** gives the longest range.
- Angles that add to 90° (like 30° and 60°) give the **same range**. The higher one stays in the air longer.
- A steeper angle gives a greater maximum height and a longer time of flight.

### Watch out: "The ball stops at the top"
Only the **vertical** velocity is zero at the top. The ball is still moving sideways at $u_x$.

### Watch out: "A heavier ball falls faster"
Without air resistance, mass does not appear in any projectile equation. A heavy ball and a light ball launched the same way follow the same path.

**Mind map**

```text
Projectile Motion
  Only gravity
    a = 9.8 m/s² down
    Path is a parabola
    Ignore air resistance
  Horizontal
    No force
    Constant velocity
    s = ut
  Vertical
    Accelerates down
    v = u + at
    Top: vertical v = 0
  Launch
    ux = u cos θ
    uy = u sin θ
    45° gives max range
  Time links both
    Same t for x and y
    Horizontal launch: t from height only
```

### Year 11: Where it comes from
These are the same equations of motion you met in Year 11 kinematics, used twice: once with $a = 0$ (horizontal) and once with $a = g$ (vertical). Velocity at any moment is the vector sum of the two parts:

$$v = \sqrt{v_x^2 + v_y^2} \qquad \tan\phi = \frac{v_y}{v_x}$$

where $\phi$ is the angle of the velocity to the horizontal.

Energy gives a shortcut for landing speed. With no air resistance, kinetic energy gained equals potential energy lost, so $v^2 = u^2 + 2gh$ for a fall of $h$, whatever the launch angle.

### Year 12: Range and maximum height formulas
For a launch and landing at the same height:

$$t = \frac{2u\sin\theta}{g} \qquad H = \frac{u^2\sin^2\theta}{2g} \qquad R = \frac{u^2\sin 2\theta}{g}$$

$\sin 2\theta$ is largest (equal to 1) when $\theta = 45°$, which is why 45° gives the maximum range. Because $\sin 2\theta = \sin(180° - 2\theta)$, the angles $\theta$ and $90° - \theta$ give the same range.

HSC questions often ask you to **compare** two launches. Write each quantity as a formula first, then see what changes. For example, doubling $u$ doubles the time of flight and multiplies the range and height by four.

When a question includes **air resistance** in words, the path is lower and shorter than the parabola, and it is no longer symmetrical: the way down is steeper than the way up.

**Reading a strobe photo.** Images taken every $\Delta t$ seconds are equally spaced sideways, so $u_x = \dfrac{\text{horizontal spacing}}{\Delta t}$. The time to the top gives $u_y = g t_{\text{top}}$.

**Landing below the launch point.** Take up as positive and use the full vertical displacement, which is negative. A stone launched at 15 m/s at 30° from a cliff lands 3.0 s later: $s_y = 7.5(3.0) - 4.9(3.0)^2 = -21.6$ m, so the cliff is 21.6 m high.

**Finding g somewhere else.** On level ground the flight time is $t = \dfrac{2u\sin\theta}{g}$, so $g = \dfrac{2u\sin\theta}{t}$.

**Which angles hit a target?** From $R = \dfrac{u^2\sin 2\theta}{g}$: with $u = 40$ m/s and $R = 120$ m, $\sin 2\theta = 0.735$, so $2\theta = 47.3°$ or $132.7°$, and $\theta = 23.7°$ or $66.3°$.

**Energy at the top.** Only $u_x$ is left, so the kinetic energy at the top is $E\cos^2\theta$, where $E$ is the launch kinetic energy. At 45° that is $\frac{E}{2}$.

**Scaling a fall.** From $h = \frac{1}{2}gt^2$, height is proportional to $t^2$: three times the fall time needs nine times the height.

**Clearing a bar or hitting a raised target.** Write $t = \dfrac{x}{u\cos\theta}$ and put it into $y = u_y t - \frac{1}{2}gt^2$. That gives the height at any distance, $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$. Set $y$ to the bar height (or target height) and solve for $u$.

**Impact angle from speeds.** At impact the horizontal speed is still $u_x$, so $\cos\phi = \dfrac{u_x}{v}$. If the speed at the top is $u$ and the impact speed is $2u$, then $\cos\phi = \frac{1}{2}$ and $\phi = 60°$ below the horizontal.

**The monkey and the hunter.** A dart aimed straight at a target that drops at the moment of firing still hits it, because both fall the same distance $\frac{1}{2}gt^2$ in the same time.

## Quick summary

- A projectile moves under gravity only. Its acceleration is 9.8 m/s² down at every point.
- Split the motion: **horizontal** at constant velocity, **vertical** with constant acceleration.
- Components of the launch velocity: $u_x = u\cos\theta$ and $u_y = u\sin\theta$.
- Time is the link between the two directions.
- At the top, vertical velocity is zero but horizontal velocity is not.
- For a horizontal launch, fall time depends only on the height.
- On level ground, 45° gives the longest range, and complementary angles give equal ranges.

## Quiz

Try these questions from the chapter quiz.

1. Ignoring air resistance, the acceleration of a projectile is:
   - A. Zero at the top of its path
   - B. Horizontal and constant
   - C. 9.8 m/s² downwards at every point
   - D. Largest at the start of the flight
   - Answer: C. 9.8 m/s² downwards at every point. Gravity is the only force, so the acceleration is g downwards all the time.

2. The horizontal velocity of a projectile (no air resistance):
   - A. Stays constant
   - B. Increases as it falls
   - C. Decreases to zero at the top
   - D. Changes direction at the top
   - Answer: A. Stays constant. There is no horizontal force, so there is no horizontal acceleration.

3. A ball is rolled off a table and another is dropped from the same height at the same time. Which lands first?
   - A. The rolled ball
   - B. The dropped ball
   - C. They land together
   - D. It depends on their masses
   - Answer: C. They land together. Both start with zero vertical velocity and fall with g, so they take the same time.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/physics/projectile-motion/#quiz