Year 12 Maths Extension 1 (2027) Vectors

Projectile motion – flight / time equations

Q: How do you find the time of flight?

Double the time to the top, giving T equals two V sin theta over g.

Q: How do you find the maximum height?

H equals V squared sin squared theta over two g, reached when the vertical velocity is zero.

Q: What angle gives the greatest range?

Forty-five degrees; complementary angles theta and ninety minus theta give the same range.

Q: How do you find the impact velocity?

Evaluate the velocity vector at the landing time; its magnitude is the impact speed.

Q: Does impact speed equal launch speed?

On level ground, yes, by symmetry of the parabola.

25 practice questions 2 video lessons Theory + worked examples

Learn the flight and time equations of projectile motion for NSW Year 12 Mathematics Extension 1. Treating a projectile's motion with vectors lets you analyse the whole flight from launch to landing.

You will learn to derive the equations of motion and find the time of flight, maximum height and range, including problems with an unknown launch speed or angle — the calculus-based projectile modelling that is a staple of the HSC Extension 1 exam.

Practice 25 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

💯 How to Find the Time of Flight for Projectile Motion Explained Watch
Projectile Motion Time of Flight Formula Derivation Watch

Theory

The flight features of a projectile: time of flight $T = \frac{2 V \sin θ}{g}$ , maximum height $H = \frac{V^{2} \sin^{2} θ}{2 g}$ , and range $R = \frac{V^{2} \sin 2 θ}{g}$ , greatest at $45^{\circ}$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

Once the components are set up, the key features of a projectile's flight — time of flight, maximum height, range and impact velocity — follow from the vertical and horizontal motion.

For launch speed $V$ at angle $θ$ from $O$ on level ground, the time to the top ( $\dot{y} = 0$ ) is $\frac{V \sin θ}{g}$ , so the time of flight is $T = \frac{2 V \sin θ}{g}$ .

Maximum height $H = \frac{V^{2} \sin^{2} θ}{2 g}$ ; range $R = \frac{V^{2} \sin 2 θ}{g}$ . On level ground the impact speed equals the launch speed.

NESA link. Part of the Year 12 Introduction to vectors focus area, outcome ME1-12-02 ("operates with 2D and 3D vectors and uses 2D vectors to solve problems involving motion in two dimensions") with MAO-WM-01.

Maximum height

H

at the apex; range

R

along the ground.

θ

and

90^{\circ} - θ

give the same range.

T = \frac{2 V \sin θ}{g}, H = \frac{V^{2} \sin^{2} θ}{2 g}, R = \frac{V^{2} \sin 2 θ}{g} .

time of flight = 2V sin(theta)/g; max height = V^2 sin^2(theta)/(2g); range = V^2 sin(2theta)/g

Key facts. Maximum height is where $\dot{y} = 0$ . Range is greatest at $θ = 45^{\circ}$ , and $θ$ and $90^{\circ} - θ$ give the same range. Flight time is twice the time to the top.

How to find flight features

Vertical velocity zero gives the time to the top; double it for the flight time.
Maximum height: the $y$ -value at the top, or $H = \frac{V^{2} \sin^{2} θ}{2 g}$ .
Range: horizontal distance at landing, or $R = \frac{V^{2} \sin 2 θ}{g}$ .
Impact velocity: $v = \dot{x} i + \dot{y} j$ at the landing time; its magnitude is the impact speed.

Example 1 — Times

A projectile is launched at

40

m/s,

θ = 30^{\circ}

(

g = 10

). Find the time to the top and the time of flight.

Solution

$V \sin θ = 40 \times \frac{1}{2} = 20$ .

$t_{top}$	$=$	$\frac{20}{10} = 2$
$T$	$=$	$2 t_{top} = 4$

time to top 2 s, flight time 4 s

Time to top $2$ s, flight time $4$ s.

Example 2 — Height and range

A projectile is launched at

20

m/s,

θ = 45^{\circ}

(

g = 10

). Find the maximum height and range.

Solution

$H$	$=$	$\frac{400 \times 0.5}{20} = 10$
$R$	$=$	$\frac{400 \sin 90^{\circ}}{10} = 40$

H = 10 m, R = 40 m

$H = 10$ m, $R = 40$ m.

Example 3 — Horizontal from a cliff

A stone is thrown horizontally at

20

m/s from an

80

m cliff (

g = 10

). Find the time to land and the impact speed.

Solution

$y = 80 - 5 t^{2} = 0 \Rightarrow t = 4$ . Then $\dot{x} = 20$ , $\dot{y} = - 40$ .

speed

=

\sqrt{20^{2} + 40^{2}} = 20 \sqrt{5}

t = 4 s, impact speed = 20 sqrt(5) m/s

$t = 4$ s, impact speed $20 \sqrt{5}$ m/s.

Example 4 — Angles for a range

20

m/s (

g = 10

), find the angles giving a range of

20 \sqrt{3}

Solution

$20 \sqrt{3}$	$=$	$\frac{400 \sin 2 θ}{10} \Rightarrow \sin 2 θ = \frac{\sqrt{3}}{2}$
$2 θ$	$=$	$60^{\circ} or 120^{\circ}$

theta = 30 or 60 degrees

So $θ = 30^{\circ}$ or $60^{\circ}$ .

Common pitfalls

Maximum height. It occurs when

\dot{y} = 0

, not when

x

is greatest.

Complementary angles. Range is greatest at

45^{\circ}

, and

θ

90^{\circ} - θ

give the same range.

Flight time. On level ground it is twice the time to the top.

Impact speed. Combine

\dot{x}

and

\dot{y}

at landing; on level ground it equals the launch speed.

Frequently asked questions

How do you find the time of flight?

Double the time to the top: $T = \frac{2 V \sin θ}{g}$ .

How do you find the maximum height?

$H = \frac{V^{2} \sin^{2} θ}{2 g}$ , reached when $\dot{y} = 0$ .

What angle gives the greatest range?

$45^{\circ}$ ; complementary angles $θ$ and $90^{\circ} - θ$ give the same range.

How do you find the impact velocity?

Evaluate $v = \dot{x} i + \dot{y} j$ at the landing time; its magnitude is the impact speed.

Does impact speed equal launch speed?

On level ground, yes — by symmetry of the parabola.

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📖 Theory

How to find flight features

Common pitfalls

Frequently asked questions

Theory