Ignoring air resistance, the acceleration of a projectile is:

9.8 m/s² downwards at every point

Largest at the start of the flight

Projectile Motion | HSC Physics Module 5

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.

Breaking the launch velocity into parts

A ball launched at speed $u$ at angle $θ$ above the horizontal has two components:

$u_{x} = u cos θ u_{y} = u sin θ$

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: $v_{y}^{2} = u_{y}^{2} + 2 a_{y} s_{y}$
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.

Launched horizontally

When something is thrown straight out sideways, $u_{y} = 0$ . The fall time depends only on the height:

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

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

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.

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 12Range and maximum height formulas

For a launch and landing at the same height:

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

$sin 2 θ$ is largest (equal to 1) when $θ = 45°$ , which is why 45° gives the maximum range. Because $sin 2 θ = sin (180° - 2 θ)$ , the angles $θ$ and $90° - θ$ 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 $Δ t$ seconds are equally spaced sideways, so $u_{x} = \frac{horizontal spacing}{Δ t}$ . The time to the top gives $u_{y} = g t_{top}$ .

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 $t = \frac{2 u sin θ}{g}$ , so $g = \frac{2 u sin θ}{t}$ .

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

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

Scaling a fall. From $h = \frac{1}{2} g t^{2}$ , 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 $t = \frac{x}{u cos θ}$ and put it into $y = u_{y} t - \frac{1}{2} g t^{2}$ . That gives the height at any distance, $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$ . 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 $cos ϕ = \frac{u _{x}}{v}$ . If the speed at the top is $u$ and the impact speed is $2 u$ , then $cos ϕ = \frac{1}{2}$ and $ϕ = 60°$ 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 $\frac{1}{2} g t^{2}$ 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: $u_{x} = u cos θ$ and $u_{y} = u sin θ$ .
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.

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Try 3 questions now

Ignoring air resistance, the acceleration of a projectile is:
1. A Zero at the top of its path
2. B Horizontal and constant
3. C 9.8 m/s² downwards at every point
4. D Largest at the start of the flight
Show answer
C: 9.8 m/s² downwards at every point
Gravity is the only force, so the acceleration is g downwards all the time.
The horizontal velocity of a projectile (no air resistance):
1. A Stays constant
2. B Increases as it falls
3. C Decreases to zero at the top
4. D Changes direction at the top
Show answer
A: Stays constant
There is no horizontal force, so there is no horizontal acceleration.
A ball is rolled off a table and another is dropped from the same height at the same time. Which lands first?
1. A The rolled ball
2. B The dropped ball
3. C They land together
4. D It depends on their masses
Show answer
C: They land together
Both start with zero vertical velocity and fall with g, so they take the same time.