# Vectors and Projectile Motion

> Model a thrown ball with vectors. Split the launch velocity into parts, find the time of flight, greatest height and range, and the path's equation.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/projectile-motion/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The perfect free kick
A soccer player kicks a ball at 20 m/s, angled $50^\circ$ above the ground. Will it clear the wall of defenders 9 m away? Will it dip under the crossbar?

Coaches and game designers answer this with one big idea: **the sideways motion and the up and down motion happen separately**. Gravity only pulls down, so it never changes how fast the ball moves sideways. Treat each direction on its own, then put them back together with vectors.

**In short:** A projectile's horizontal and vertical motions are separate, and gravity only affects the vertical one. Launched at speed $u$ and angle $\theta$, $x = ut\cos\theta$ and $y = ut\sin\theta - \dfrac{1}{2}gt^2$, and on flat ground the range is greatest at $45^\circ$.

## Starting point: what you already know

- Vectors in $\mathbf{i}$, $\mathbf{j}$ form, and $r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}$ for a vector of length $r$ at angle $\theta$.
- From Mathematics Advanced: velocity is the derivative of displacement, $v = \dfrac{dx}{dt}$, and acceleration is the derivative of velocity. Going backwards uses integration.
- Gravity gives every falling object an acceleration of about $g = 9.8$ m/s² downward (often rounded to 10 in exam questions).
- Solving quadratics and finding the maximum of a parabola.

## Setting up the vectors

Put the origin at the launch point, $\mathbf{i}$ horizontal and $\mathbf{j}$ up. The only force (ignoring air resistance) is gravity, so

$$\mathbf{a}(t) = -g\,\mathbf{j}$$

Launch with speed $u$ at angle $\theta$ to the horizontal:

$$\mathbf{v}(0) = u\cos\theta\,\mathbf{i} + u\sin\theta\,\mathbf{j}$$

%%FIG:projectile%%

Integrate once for velocity and again for position (the constants come from the starting values):

### Key idea: The equations of motion
$$\mathbf{v}(t) = u\cos\theta\,\mathbf{i} + (u\sin\theta - gt)\,\mathbf{j}$$
$$\mathbf{r}(t) = ut\cos\theta\,\mathbf{i} + \left(ut\sin\theta - \tfrac{1}{2}gt^2\right)\mathbf{j}$$
So $x = ut\cos\theta$ and $y = ut\sin\theta - \tfrac{1}{2}gt^2$.

Notice that the horizontal velocity $u\cos\theta$ never changes.

### Calculator tip: Store the components
Work out $u\cos\theta$ and $u\sin\theta$ once and store them in `A` and `B` with `STO`. Then the time of flight, greatest height and range use `A` and `B` without rounding. Check you are in degree mode when $\theta$ is in degrees.

## The three big questions

Use the free kick: $u = 20$, $\theta = 50^\circ$, $g = 10$. Then $u\cos\theta \approx 12.86$ and $u\sin\theta \approx 15.32$.

**1. Greatest height.** At the top, the ball stops going up: vertical velocity is 0.
$$15.32 - 10t = 0 \quad\Rightarrow\quad t \approx 1.53 \text{ s}$$
$$y = 15.32(1.53) - 5(1.53)^2 \approx 11.7 \text{ m}$$

**2. Time of flight.** It lands when $y = 0$:
$$t(15.32 - 5t) = 0 \quad\Rightarrow\quad t \approx 3.06 \text{ s}$$
This is twice the time to the top, as symmetry suggests.

**3. Range.** Horizontal distance at landing:
$$x = 12.86 \times 3.06 \approx 39.4 \text{ m}$$

### Worked example: Does it clear the wall?
The wall is at $x = 9$. Time to get there: $t = \dfrac{9}{12.86} \approx 0.70$ s. Height then: $y = 15.32(0.70) - 5(0.70)^2 \approx 8.3$ m. The wall is about 2 m high, so the ball clears it easily.

### Exam tip: Speed and direction at any time
Speed is the magnitude of $\mathbf{v}(t)$. The direction of motion makes angle $\alpha$ with the horizontal where $\tan\alpha = \dfrac{\text{vertical velocity}}{\text{horizontal velocity}}$.

**Mind map**

```text
Projectile motion
  Set up
    Origin at launch
    a = -g j
    v(0) = u cos θ i + u sin θ j
  Equations
    x = ut cos θ
    y = ut sin θ - ½gt²
  Questions
    Top: vertical velocity 0
    Flight time: y = 0
    Range: x at landing
  Path
    Remove t
    A parabola
```

### Year 11: The path is a parabola
From $x = ut\cos\theta$, $t = \dfrac{x}{u\cos\theta}$. Substitute into $y$:
$$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} = x\tan\theta - \frac{gx^2}{2u^2}\left(1 + \tan^2\theta\right)$$
This is the Cartesian equation of the path, a downward parabola. It is the parametric idea from Year 11, with time as the parameter.

### Year 12: General formulas and the best angle
Doing the same steps with letters:
$$\text{time of flight } T = \frac{2u\sin\theta}{g}, \quad \text{greatest height } H = \frac{u^2\sin^2\theta}{2g}, \quad \text{range } R = \frac{u^2\sin 2\theta}{g}$$
Since $\sin 2\theta$ is largest when $2\theta = 90^\circ$, the longest range on flat ground comes from $\theta = 45^\circ$. Angles that add to $90^\circ$, such as $30^\circ$ and $60^\circ$, give the same range.

**Hitting a target on level ground.** Set the range equal to the distance: $\dfrac{u^2\sin 2\theta}{g} = d$. With $u = 20$, $g = 10$ and $d = 20\sqrt{3}$: $\sin 2\theta = \frac{\sqrt{3}}{2}$, so $2\theta = 60^\circ$ or $120^\circ$, and $\theta = 30^\circ$ or $60^\circ$.

**Passing through a point.** Use the Cartesian path with $\dfrac{1}{\cos^2\theta} = 1 + \tan^2\theta$, which gives a quadratic in $T = \tan\theta$. To pass through $(10, 5)$ with $u = 20$ and $g = 10$: $5 = 10T - \frac{10 \times 100}{800}(1 + T^2)$, so $1.25T^2 - 10T + 6.25 = 0$, that is $T^2 - 8T + 5 = 0$ and $T = 4 \pm \sqrt{11}$. Two angles work.

**Speed at the top.** At the top only $u\cos\theta$ is left. If that is half the launch speed, $\cos\theta = \frac{1}{2}$ and $\theta = 60^\circ$.

**Collisions.** Two projectiles launched at the same moment from the same height collide only if they are always at the same height, so their vertical launch speeds must be equal: $u_1\sin\alpha = u_2\sin\beta$.

**Up a slope.** Given a formula such as the greatest range up a slope $\dfrac{u^2}{g(1 + \sin\alpha)}$, just substitute. Using the formula correctly is the skill being tested.

### Year 12: Launching from a height
A ball is thrown horizontally at 15 m/s from a cliff 45 m high. Take the origin at the foot of the cliff, so $y = 45 - 5t^2$ and $x = 15t$. It lands when $y = 0$: $t = 3$ s, at $x = 45$ m from the cliff.

**Speed on landing.** At landing $t = 3$, the velocity is $15\mathbf{i} - 30\mathbf{j}$, so the speed is $\sqrt{225 + 900} = 15\sqrt{5} \approx 33.5$ m/s, at $\tan^{-1}2 \approx 63^\circ$ below the horizontal.

## Quick summary

- Horizontal and vertical motion are separate. Gravity only affects the vertical part.
- $x = ut\cos\theta$ and $y = ut\sin\theta - \tfrac{1}{2}gt^2$.
- Greatest height: vertical velocity is 0. Flight time: $y = 0$. Range: $x$ at landing.
- On flat ground the range is $\dfrac{u^2\sin 2\theta}{g}$, greatest at $45^\circ$.

## Quiz

Try these questions from the chapter quiz.

1. Ignoring air resistance, the acceleration of a projectile is:
   - A. zero
   - B. $g$ downward
   - C. $g$ forward
   - D. it changes with time
   - Answer: B. $g$ downward. Only gravity acts.

2. Which part of the velocity stays the same during flight?
   - A. the horizontal part
   - B. the vertical part
   - C. both parts
   - D. neither part
   - Answer: A. the horizontal part. Gravity has no horizontal effect.

3. A ball is launched at 10 m/s at $30^\circ$. Its initial vertical velocity is:
   - A. 10 m/s
   - B. 5 m/s
   - C. $5\sqrt{3}$ m/s
   - D. 0 m/s
   - Answer: B. 5 m/s. $10\sin 30^\circ = 5$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/projectile-motion/#quiz