Projectile Motion

Why a projectile's horizontal and vertical motions can be treated as two separate problems, and how to solve both of them under exam conditions.

55 min · Module 5: Advanced Mechanics

The full teaching. Start here.

Syllabus requirementsModule 5: Advanced Mechanics

analyse the motion of projectiles by resolving the motion into horizontal and vertical components, making the following assumptions:
a constant vertical acceleration due to gravity
zero air resistance
apply the modelling of projectile motion to quantitatively derive the relationships between the following variables:
initial velocity
launch angle
maximum height
time of flight
final velocity
launch height
horizontal range of the projectile (ACSPH099)ACSPH099
conduct a practical investigation to collect primary data in order to validate the relationships derived above.
solve problems, create models and make quantitative predictions by applying the equations of motion relationships for uniformly accelerated and constant rectilinear motion

What this actually asks for

Three things, and the order matters.

The first asks you to resolve: to stop treating the curved path as one motion and start treating it as two straight line motions happening at once. The two assumptions are named as dot points of their own, which means they can be asked about directly, not just relied on.

The second says derive, not apply. Seven quantities are listed, and the examination can ask you to produce the relationship between any of them from the components rather than quote a remembered formula.

The third and fourth are the practical and the problem solving. You need to be able to collect data that could disagree with the model, and to solve for any unknown when the question does not hand you the time.

The short version

A projectile is anything moving under gravity alone. Once it has left the hand, the bat or the ramp, one force acts on it: its weight, pointing straight down.

That single fact is the whole topic. Gravity is vertical, so it can only change vertical velocity. The horizontal velocity is left alone for the entire flight.

So split every projectile question into two columns and never let a number cross between them.

Horizontal. Acceleration is zero. Velocity stays at $u_{x} = u cos θ$ from launch to landing. Distance is just $x = u_{x} t$ .

Vertical. Acceleration is $9.8$ m s⁻² downwards, always. Use the three equations of motion with $u_{y} = u sin θ$ .

The two columns share exactly one quantity: the time. Time is the bridge, and almost every projectile question is solved by finding $t$ in one column and carrying it to the other.

Four results worth having in your head, all for a launch from level ground:

Time to the top is $u_{y} / g$ , and the flight is symmetric, so the total time is twice that.
Greatest height above the launch point is $u_{y}^{2} /2 g$ .
Range is $R = u^{2} sin 2 θ / g$ , only when the landing height equals the launch height.
The landing speed equals the launch speed, and the landing angle equals the launch angle.

The fastest way to lose marks here is to use result 3 on a launch from a cliff, a bench or a shoulder. Check the heights before you reach for it.

One force, two problems

Watch a thrown ball and you see a single smooth curve. That curve is hard to think about directly. There is no equation in the course for a parabola travelled at a changing speed, and there does not need to be, because the curve is not the thing that is happening. It is the shadow of two much simpler things happening together.

Why the horizontal and vertical motions do not affect each other

AnswersA ball is moving sideways and falling at the same time. Why does the sideways part not slow the falling down, or the falling not curve the sideways part?

Once a projectile has left whatever launched it, exactly one force acts on it: its weight, $m g$ , pointing straight down. In this topic we are told to neglect air resistance, so there is nothing else.

Newton's second law says the acceleration points the same way as the net force. The net force points straight down. So the acceleration points straight down, and has no horizontal part at all.

An acceleration with no horizontal part cannot change a horizontal velocity. Not slowly, not a little. The horizontal velocity a projectile has at launch is the horizontal velocity it has when it lands, and at every instant in between.

Meanwhile the vertical velocity changes at a steady $9.8$ m s⁻² downwards, because the vertical acceleration is steady. Whether the object is also travelling sideways at $2$ m s⁻¹ or $200$ m s⁻¹ appears nowhere in that sentence, because horizontal velocity is not one of the things gravity responds to.

So we have two motions that share a clock and share nothing else. Each one is a problem you already know how to solve: uniform velocity in one direction, uniform acceleration in the other.

The classic demonstration is two balls leaving a bench at the same instant, one dropped and one fired horizontally. They hit the floor together. It looks wrong the first time you see it, and it is the most useful thing in the module.

Speed

Launch speed25 m s⁻¹

Launch angle50 °

Launch height0 m

Path of the projectile

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

The numbers behind this graph

Time (s)	Horizontal distance (m)	Height (m)
0.00	0.0	0.0
0.78	12.6	12.0
1.56	25.1	18.0
2.35	37.7	18.0
3.13	50.2	12.0
3.91	62.8	0.0

Watch the horizontal spacing of the dots. It never changes, from launch to landing, no matter what angle you choose.

The dots are dropped at equal time intervals. Look at their horizontal spacing: it never changes, from launch to landing, whatever angle you choose. Now look at their vertical spacing: close together near the top, further apart near the ground. Two motions, one clock.

CheckpointAnswer before reading on.

Two identical steel balls leave the edge of a bench at the same instant. Ball A is simply released. Ball B is fired horizontally at $4.0$ m s⁻¹. Air resistance is negligible.

Which statement about the two balls is correct?

AThey strike the floor at the same instant, because each starts with zero vertical velocity and each accelerates downwards at $9.8$ m s⁻².
BBall B lands later, because its path is longer.
CBall B lands earlier, because it is moving faster.
DBall A lands earlier, because the whole of its motion is downwards.

Hint 1Split each ball's motion into a horizontal part and a vertical part before comparing them.

Hint 2Gravity pulls straight down, so it can only change the vertical part of a velocity.

Hint 3Write down the vertical starting velocity of each ball. If those agree, and the vertical acceleration agrees, the vertical motions are the same motion.

The two assumptions, stated out loud

The syllabus names these as dot points in their own right, which tells you they can be asked about directly.

Projectile: An object moving through the air with no propulsion of its own, so that the only force acting on it is gravity. A thrown ball is a projectile. A rocket under thrust is not. A ball on a string is not, because the string pulls on it.

Assumption one: a constant vertical acceleration due to gravity. We take $g = 9.8$ m s⁻² downwards and treat it as the same everywhere in the flight. This is a good approximation near the Earth's surface, where $g$ changes by about $0.003$ per cent over a hundred metres of height. It fails for a flight long enough to matter, which is why an intercontinental missile is not a projectile motion problem.

Assumption two: zero air resistance. This is the bigger lie of the two, and the one you can be asked to assess. It matters more the faster, lighter and larger the object is. A shot put is described well. A shuttlecock is described so badly that its real path looks nothing like a parabola.

Both assumptions are what make the two directions independent. Air resistance acts along the direction of motion, which is neither horizontal nor vertical, so it would couple the two columns together and the method would collapse.

Resolving the launch velocity

Every projectile answer begins the same way, before any equation of motion is written: take the launch velocity apart.

Resolving a launch velocity into componentsnot on the formulae sheet, learn it

u_{x} = u cos θ u_{y} = u sin θ

Symbols: $u$ launch speed, the magnitude of the launch velocitym s⁻¹
$θ$ launch angle, measured from the horizontaldegrees or radians
$u_{x}$ horizontal component, unchanged for the whole flightm s⁻¹
$u_{y}$ vertical component at launchm s⁻¹
Valid when: The launch angle is measured from the horizontal, which is the convention every HSC question uses unless it says otherwise.
Not valid when: The angle is given from the vertical, from a slope, or from a line of sight. Then the sine and cosine swap, or the angle has to be converted before either is used.
Rearranged: for u: $u = u_{x}^{2} + u_{y}^{2}$ for \theta: $θ = tan^{- 1} (\frac{u _{y}}{u _{x}})$
Where it turns up: The first line of almost every projectile answer, before any equation of motion is written
Recombining a final horizontal and vertical velocity into a landing speed and a landing angle
Checking that a component is smaller than the speed it came from, which catches a calculator left in radians
Where marks go missing: Swapping sine and cosine, which is caught instantly by asking whether a shallow launch should have a large or a small horizontal component
Leaving the calculator in radian mode, which produces components larger than the launch speed
Resolving once and then forgetting to keep the horizontal component constant for the rest of the flight

Where this comes from

One flight, six things the examination asks about

1Launch: resolve u into components
Nothing happens until the launch velocity is split. A speed of $u$ at an angle $θ$ above the horizontal becomes $u_{x} = u cos θ$ going sideways and $u_{y} = u sin θ$ going up. From this point on, $u$ and $θ$ are rarely used again.
2On the way up: vertical velocity shrinking
The vertical velocity falls by $9.8$ m s⁻¹ every second. The horizontal velocity does not change. The path leans over because one of the two arrows is shrinking while the other stays the same length.
3Greatest height: vertical velocity zero
The vertical velocity is zero here, which is what greatest height means. The velocity itself is not zero: the projectile is still moving horizontally at $u_{x}$ . The acceleration is not zero either. It is still $9.8$ m s⁻² downwards, because the weight has not gone anywhere.
4On the way down: mirror image of the rise
Everything now runs in reverse. At a given height on the way down the speed matches the speed at that same height on the way up, and the time from the peak to any height matches the time from that height to the peak. This symmetry holds only because the launch and landing heights are equal.
5Landing: same speed, same angle, downwards
On level ground the projectile arrives at the speed it left with, at the same angle below the horizontal as it was launched above it. Its horizontal component was never touched, and its vertical component has been reversed exactly.
6Range: horizontal speed times total time
The range is not a separate physical quantity. It is the horizontal speed multiplied by the total time, and the total time came from the vertical column. Every range calculation is really a vertical calculation with one multiplication at the end.

The same parabola that a level ground launch produces, with the quantities named in the syllabus marked on it.

The components are worth sanity checking every time. Both of them must be smaller than the launch speed, since a right angled triangle's legs are shorter than its hypotenuse. If a component comes out larger, the calculator is in radian mode.

Horizontal velocity, constant
Vertical velocity, falling at 9.8 m s⁻²
Speed, the two combined

This is the whole idea in one picture. Gravity acts vertically, so only the vertical line has a gradient. The flat line is why horizontal distance is just speed times time.

Launch speed25 m s⁻¹

Launch angle45 °

Horizontal component: 17.68 m s⁻¹, unchanging
Vertical component at launch: 17.68 m s⁻¹ up
Vertical component at landing: 17.68 m s⁻¹ down
Time to the top: 1.80 s

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

That graph is the argument of this lesson in one picture. The flat line is the horizontal component, which has no gradient because it has no acceleration. The sloping line is the vertical component, falling at $9.8$ m s⁻¹ every second and passing through zero at the top of the flight. There is no kink where they cross zero, because nothing happens to the force there.

CheckpointAnswer before reading on.

A ball is launched at $30$ m s⁻¹, $40°$ above the horizontal. Air resistance is negligible.

What is the horizontal component of its velocity $1.5$ s after launch?

A $23$ m s⁻¹
B $8.3$ m s⁻¹
C $30$ m s⁻¹
D $19$ m s⁻¹

Hint 1Work out the horizontal component at launch first, then ask what could possibly have changed it.

Hint 2The only force acting is gravity, and gravity points straight down.

Hint 3 $u_{x} = u cos θ$ , and the horizontal acceleration is zero for the entire flight.

CheckpointAnswer before reading on.

A projectile is launched at an angle over level ground. Which statement describes it correctly at the highest point of its path?

AThe velocity is horizontal, and the acceleration is $9.8$ m s⁻² downwards.
BBoth the velocity and the acceleration are zero.
CThe velocity is zero, and the acceleration is $9.8$ m s⁻² downwards.
DThe velocity is horizontal, and the acceleration is zero.

Hint 1Ask about the two components separately. One of them is doing something interesting at the top, the other is not.

Hint 2Acceleration is set by the forces acting, not by how fast the object happens to be going.

Hint 3At the top the vertical velocity passes through zero. The weight is still the only force acting, and it has not changed.

CheckpointAnswer before reading on.

A projectile is launched at $25$ m s⁻¹, $60°$ above the horizontal, from level ground. Taking upwards as positive, which description matches a graph of the vertical component of its velocity against time for the whole flight?

AA single straight line starting at $+ 21.7$ m s⁻¹, sloping down at a constant rate, crossing zero halfway through the flight and finishing at $- 21.7$ m s⁻¹.
BA straight line falling to zero at the top of the flight, then a second straight line rising back to $+ 21.7$ m s⁻¹.
CA curve starting steep and flattening out, approaching zero at the end of the flight.
DA horizontal line at $+ 12.5$ m s⁻¹ for the whole flight.

Hint 1The gradient of a velocity against time graph is the acceleration.

Hint 2The vertical acceleration is constant at $9.8$ m s⁻² downwards for every instant of the flight, including the top.

Hint 3Work out the vertical velocity at launch and at landing, then join them with a line of the right gradient.

The method, and the three equations

Horizontally there is nothing to learn. The acceleration is zero, so $s = u t + \frac{1}{2} a t^{2}$ collapses to

$x = u_{x} t$

Vertically you have the full set, all three on the NESA formulae sheet.

Displacement under uniform accelerationon the NESA formulae sheet

s = u t + \frac{1}{2} a t^{2}

Symbols: $s$ displacement over the interval, not distance travelledm
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$t$ length of the intervals
Valid when: The acceleration is constant and the displacement, initial velocity and acceleration are all measured along the same line in the same positive direction.
Not valid when: The acceleration varies, or the answer wanted is total path length rather than displacement. A projectile thrown up and caught again has zero vertical displacement while having travelled a real distance.

Velocity after uniform accelerationon the NESA formulae sheet

v = u + a t

Symbols: $v$ velocity at the end of the intervalm s⁻¹
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$t$ length of the intervals
Valid when: The acceleration is constant and every symbol refers to the same straight line, with one direction chosen as positive and kept that way.
Not valid when: The acceleration changes during the interval, or symbols from two different directions are mixed. In projectile work this equation belongs to the vertical direction only, because the horizontal acceleration is zero.

Velocity against displacement, with no timeon the NESA formulae sheet

v^{2} = u^{2} + 2 a s

Symbols: $v$ velocity at the end of the intervalm s⁻¹
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$s$ displacement over the intervalm
Valid when: The acceleration is constant along the line of motion and the question gives or wants a displacement rather than a time.
Not valid when: The acceleration changes, or the question needs a time. It also loses the sign of the answer, since squaring destroys direction, so the direction has to be reasoned back in from the situation.

Choosing which one to use is a matter of asking what the question gives you and what it wants. If you have a time or want one, the first two are candidates. If the question mentions no time and wants none, the third saves you a step.

Write your positive direction down before you write anything else

Half the sign errors in this topic come from deciding, silently and separately, what up means in each line of working. Put taking up as positive at the top of the answer and then give every quantity a sign that agrees with it: $g$ becomes $- 9.8$ , a downward displacement becomes negative, a landing point below the launch point is a negative $s$ .

Markers award method marks for a correct setup even when the arithmetic later goes wrong, and a stated sign convention is part of that setup.

The simplest case: launched horizontally

Worked example4 marks

A marble rolls off a bench

A marble rolls off the edge of a bench $0.92$ m high, travelling horizontally at $1.8$ m s⁻¹. Find the time it takes to reach the floor, how far from the bench it lands, and its speed and direction on impact. Take $g = 9.8$ m s⁻² and neglect air resistance.

Set up the two columns

Doing this first means no number ever ends up in the wrong equation, which is where most of the marks in this topic are lost.

Take down as positive, since everything of interest happens downwards.

Horizontally: $u_{x} = 1.8$ m s⁻¹, $a_{x} = 0$ .

Vertically: $u_{y} = 0$ , $a_{y} = 9.8$ m s⁻², $s_{y} = 0.92$ m.

The phrase rolls off horizontally is what sets $u_{y} = 0$ . The marble is not thrown downwards, so it begins the fall from rest in the vertical direction.

Find the time, which has to come from the vertical column

The horizontal column contains two unknowns, the distance and the time, so it cannot be solved on its own. The vertical column contains only one.

$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} = 0 + \frac{1}{2} (9.8) t^{2}$

$0.92 = 4.9 t^{2} \Rightarrow t^{2} = 0.1878 \Rightarrow t = 0.433 s$

Carry the time across to the horizontal column

Time is the only quantity the two columns share. This is the step the whole method exists for.

$x = u_{x} t = 1.8 \times 0.433 = 0.78 m$

Build the impact velocity from its two components

Speed and direction are properties of the combined velocity, so the components have to be recombined at the very end, never earlier.

The horizontal component is unchanged at $1.8$ m s⁻¹. The vertical component has grown from zero:

$v_{y} = u_{y} + a_{y} t = 0 + 9.8 \times 0.433 = 4.25 m s^{- 1}$

$v = v_{x}^{2} + v_{y}^{2} = 1. 8^{2} + 4.2 5^{2} = 4.6 m s^{- 1}$

$α = tan^{- 1} (\frac{4.25}{1.8}) = 67°$

Answer

The marble is in the air for $0.43$ s, lands $0.78$ m from the base of the bench, and strikes the floor at $4.6$ m s⁻¹ at $67°$ below the horizontal.

Is that answer sensible?

Bench height falls take about half a second, which matches. The marble lands less than a metre out, which is what a slow roll should give. The impact is much steeper than it is fast sideways, and that fits: the vertical component grew to more than twice the horizontal one during the fall, so the velocity should point mostly downwards.

CheckpointAnswer before reading on.

A stone is thrown horizontally at $12$ m s⁻¹ from the top of a cliff $45$ m above the sea.

How long does the stone take to reach the water? Take $g = 9.8$ m s⁻² and neglect air resistance.

Your answers

Give it to 3 significant figures.

Hint 1Time is a vertical question here. Decide which of the two directions actually contains the information you need.

Hint 2Thrown horizontally means the vertical component of the launch velocity is zero.

Hint 3Take down as positive and use $s = u t + \frac{1}{2} a t^{2}$ vertically, with $u = 0$ , $a = 9.8$ and $s = 45$ .

The general case: launched at an angle

Worked example7 marks

A ball kicked over level ground

A ball is kicked from level ground at $22$ m s⁻¹, $38°$ above the horizontal, and lands on the same level ground. Find the greatest height it reaches, the time of flight, the horizontal range, and its velocity on landing. Take $g = 9.8$ m s⁻² and neglect air resistance.

Resolve, and state the sign convention

Every later line uses these two numbers. Getting them onto the page once stops the launch speed being substituted into a vertical equation by mistake.

Taking up as positive, so $a_{y} = - 9.8$ m s⁻².

$u_{x} = 22 cos 38° = 17.3 m s^{- 1} u_{y} = 22 sin 38° = 13.5 m s^{- 1}$

Both are smaller than $22$ , as they must be.

Greatest height, from the condition that defines it

Greatest height is not a formula to memorise. It is the height at the one instant when the vertical velocity is zero, and that condition turns an equation of motion into an answer.

At the top, $v_{y} = 0$ . No time is given and none is wanted, so use the equation without $t$ in it:

$v_{y}^{2} = u_{y}^{2} + 2 a_{y} s_{y} \Rightarrow 0 = 13. 5^{2} + 2 (- 9.8) s_{y}$

$s_{y} = \frac{13. 5 ^{2}}{19.6} = 9.3 m$

Time of flight, from the vertical displacement over the whole flight

The landing point is the same height as the launch point, so the vertical displacement for the entire flight is zero. That single substitution is what makes the level ground case easy.

$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow 0 = 13.5 t - 4.9 t^{2} = t (13.5 - 4.9 t)$

$t = 0 or t = \frac{13.5}{4.9} = 2.76 s$

The root $t = 0$ is the launch itself, so the time of flight is $2.76$ s. Notice it is exactly twice the time to the top, $13.5/9.8 = 1.38$ s. That symmetry is a property of the equal heights, not of projectiles in general.

Range, by carrying the time to the horizontal column

Horizontal motion is uniform, so this is the one genuinely trivial step in the question.

$x = u_{x} t = 17.3 \times 2.76 = 48 m$

Landing velocity, by recombining the components

The two components are tracked separately for the whole flight and only ever put back together at the moment a question asks for a speed or a direction.

Horizontally, still $17.3$ m s⁻¹.

$v_{y} = u_{y} + a_{y} t = 13.5 - 9.8 \times 2.76 = - 13.5 m s^{- 1}$

$v = 17. 3^{2} + 13. 5^{2} = 22 m s^{- 1} α = tan^{- 1} (\frac{13.5}{17.3}) = 38°$

Answer

Greatest height $9.3$ m, time of flight $2.76$ s, range $48$ m, landing at $22$ m s⁻¹ at $38°$ below the horizontal.

Is that answer sensible?

The landing speed came out equal to the launch speed and the landing angle equal to the launch angle. That is not a coincidence and it is a free check on the whole answer: on level ground with no air resistance, the flight is symmetric, so those two pairs must match. If yours do not, the error is upstream.

Sizes are sensible too. A well struck ball travelling about fifty metres and staying up for not quite three seconds is what a football looks like.

Deriving the range relationship

The syllabus asks you to derive the relationships between the listed variables, not to recall them. Here is the one that is asked for most often, produced from the two columns rather than quoted.

Derivation

Range on level ground

Starts from

the two component equations, with the vertical displacement over the whole flight set to zero

Ends at

R = \frac{u ^{2} sin 2 θ}{g}

Holds only if

The projectile lands at exactly the height it was launched from
Air resistance is negligible
The launch angle is measured from the horizontal
g is constant throughout the flight

Range on level ground

5 steps

1
Resolve the launch velocity
Nothing in the derivation can start until the single launch velocity has become two independent ones.
$u_{x} = u cos θ u_{y} = u sin θ$
2
Write the vertical displacement over the whole flight
This is the only place the level ground condition enters, and it is the assumption that later gets forgotten.
Taking up as positive, the projectile finishes at the height it started, so $s_{y} = 0$ over the whole flight:
$0 = u sin θ t - \frac{1}{2} g t^{2}$
3
Solve for the time of flight
Factorising rather than using the quadratic formula makes the two roots and their meanings visible.
$0 = t (u sin θ - \frac{1}{2} g t)$
The root $t = 0$ is the launch. The one we want is
$t = \frac{2 u s i n θ}{g}$
4
Carry the time into the horizontal equation
Range is a horizontal distance, and the horizontal column has been waiting for a time the whole way through.
$R = u_{x} t = u cos θ \times \frac{2 u s i n θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g}$
5
Tidy with the double angle identity
The compressed form is what makes the two properties of the range obvious, which is the reason the identity is worth applying rather than leaving the expression as it is.
Since $2 sin θ cos θ = sin 2 θ$ ,
$R = \frac{u ^{2} s i n 2 θ}{g}$
Two things now fall out for free. The sine reaches its largest value when $2 θ = 90°$ , so the greatest range comes at $θ = 45°$ . And because $sin x = sin (180° - x)$ , angles that add to $90°$ give the same range.

Range of a projectile on level groundnot on the formulae sheet, learn it

R = \frac{u ^{2} sin 2 θ}{g}

Symbols: $R$ horizontal distance from launch to landingm
$u$ launch speedm s⁻¹
$θ$ launch angle above the horizontaldegrees or radians
$g$ acceleration due to gravity, 9.8 on the NESA data sheetm s⁻²
Valid when: The projectile lands at exactly the height it was launched from, air resistance is ignored, and the angle is measured from the horizontal.
Not valid when: The launch and landing heights differ by any amount. A ball thrown from a cliff, a shot released at shoulder height, or a serve struck above the far court all break it, and in each case the components have to be worked through instead.
Rearranged: for u: $u = \frac{R g}{s i n 2 θ}$ for \theta: $θ = \frac{1}{2} sin^{- 1} (\frac{R g}{u ^{2}})$
Where it turns up: Showing that the greatest range on level ground comes at 45 degrees, where the sine reaches one
Showing that two launch angles adding to 90 degrees give the same range, because they give the same doubled angle
A fast check on an answer that was properly worked through with components
Where marks go missing: Using it when the launch height and the landing height are different, which is the single most common way marks are lost in this topic
Writing sine of theta rather than sine of two theta, which gives the wrong answer at every angle except the one where they happen to agree
Quoting it in an examination that expects the relationships to be derived, where the marks are for the components and not for a remembered result

Where this comes from

Range achieved

Two angles give the same range, and that is not a coincidence. Raise the launch point and watch the peak slide left of 45 degrees.

Launch speed25 m s⁻¹

Launch height0 m

Best angle: 45.0°
Greatest range: 63.8 m
Range at 45°: 63.8 m

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Drag the launch height slider away from zero and watch the peak of the curve slide to the left of $45°$ . That is the assumption in the derivation making itself felt: the moment the landing height stops matching the launch height, the result stops holding.

CheckpointAnswer before reading on.

A projectile launched from level ground at $25°$ above the horizontal travels a range $R$ . Air resistance is negligible.

Launched at the same speed, which other angle gives the same range $R$ ?

A $65°$
B $50°$
C $45°$
D $75°$

Hint 1Write the range in terms of the launch angle rather than reasoning about the picture.

Hint 2The range on level ground depends on $sin 2 θ$ , not on $sin θ$ .

Hint 3Two different angles between $0°$ and $90°$ give the same sine when the doubled angles add to $180°$ .

Common mistake

Using R = u²sin2θ/g for a launch from a cliff, a bench, a shoulder or a racquet above the net.

Why it happens: The equation is the only range result most students have memorised, and the question always supplies a launch speed and an angle, so it looks like it fits.

The vertical displacement was set to zero in the third line of the derivation. That step encoded an assumption which the equation itself no longer displays, and an assumption you cannot see is an assumption you will not check.

The tell is in the question, not in the formula. If the launch point and the landing point are at different heights by any amount, go back to the components: find the time of flight from the vertical equation with the real displacement, then multiply by $u_{x}$ .

In the cliff question later in this lesson the level ground result would give $31$ m against a true answer of $52$ m. It is not a small error and it does not look like one from the inside.

Common mistake

Setting the vertical displacement to +25 m for a projectile that lands 25 m below its launch point.

Why it happens: The number appears in the question as a positive height above the ground, so it gets carried into the equation with the sign it was printed with.

A displacement is measured from the launch point, in the direction you nominated as positive. With up positive, a landing point below the launch point is $s_{y} = - 25$ m.

The practical defence is to write the sign convention at the top and then, for each quantity in turn, ask which way it points rather than what number the question printed.

When the question gives you no time

This is the step that separates a confident answer from a stalled one. Some questions specify a point the projectile must pass through, and ask for the launch speed or the launch angle. Time appears in both columns and is given in neither.

The move is to eliminate it: solve the horizontal equation for $t$ and substitute into the vertical one.

Worked example4 marks

Does the ball clear the wall?

A ball is kicked from level ground at $25$ m s⁻¹, $30°$ above the horizontal. A wall $4.0$ m high stands $40$ m away. Determine whether the ball passes over the wall, and by how much. Take $g = 9.8$ m s⁻² and neglect air resistance.

Turn the question into a question about one instant

A wall is not a landing point. What is being asked is the height of the ball at the moment its horizontal distance is 40 m, which is a much more concrete thing to calculate.

Taking up as positive:

$u_{x} = 25 cos 30° = 21.7 m s^{- 1} u_{y} = 25 sin 30° = 12.5 m s^{- 1}$

Use the horizontal column to find when the ball is level with the wall

The horizontal column is the one with no acceleration, so it converts a distance into a time in a single division.

$t = \frac{x}{u _{x}} = \frac{40}{21.7} = 1.85 s$

Use that time in the vertical column to find the height there

This is the same carrying of the time between columns as before, just run in the opposite direction: horizontal first, vertical second.

$s_{y} = u_{y} t - \frac{1}{2} g t^{2} = 12.5 (1.85) - 4.9 (1.85)^{2}$

$s_{y} = 23.1 - 16.8 = 6.3 m$

Compare, and say which side of the wall the answer falls on

A number on its own does not answer a yes or no question. The final mark in a question phrased like this is for the comparison and the conclusion.

The wall is $4.0$ m high and the ball is at $6.3$ m when it arrives there, so it passes over with $2.3$ m to spare.

Answer

The ball clears the wall by $2.3$ m.

Is that answer sensible?

Is the ball still rising at that point? Time to the top is $12.5/9.8 = 1.28$ s, and the ball reaches the wall at $1.85$ s, so it is on the way down. That fits: the greatest height is $12. 5^{2} /19.6 = 8.0$ m, and $6.3$ m is below that, which is consistent with a ball that has already peaked.

Exam question

Harder · about 9 min

6 marks

A ball is projected from the edge of a cliff at $18$ m s⁻¹, $35°$ above the horizontal. The launch point is $25$ m above level ground, and the ball lands on that ground. Take $g = 9.8$ m s⁻² and neglect air resistance.

(a) Calculate the greatest height of the ball above the ground. (2 marks)

(b) Calculate the time of flight. (2 marks)

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Resolve the launch velocity once, at the top of the page, and use those two numbers for all three parts.

Hint 2Greatest height is where the vertical velocity is zero. Time of flight is where the vertical displacement equals $- 25$ m. Horizontal distance is the horizontal speed times that time.

Hint 3For (b), taking up as positive gives $- 25 = u_{y} t - 4.9 t^{2}$ . Rearrange to $4.9 t^{2} - u_{y} t - 25 = 0$ and use the quadratic formula, keeping the positive root.

Where students lose marks on this one

Answering $5.4$ m for part (a).
Why it happens: The equation of motion returns a displacement measured from the launch point, and the question asks for a height measured from the ground. The calculation is right and the answer to the question asked is not.
Underline the reference level in the question before writing anything. Here the phrase is above the ground, so the $25$ m has to be added.
Using $R=u^{2}\sin 2\theta/g$ for part (c).
Why it happens: It is the fastest route to a range and it is remembered as *the* range formula.
That result assumes the projectile lands at its launch height. Here it lands $25$ m lower, so it would give $31$ m instead of $52$ m, an error of twenty one metres on a fifty metre answer.
Setting the vertical displacement to $+25$ m in part (b).
Why it happens: The number $25$ appears in the question as a positive height, so it gets substituted with the sign it was printed with.
Choose a positive direction, write it down, and then give every quantity a sign consistent with it. With up positive, a landing point below the launch point is a negative displacement.

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Exam question

Hardest asked · about 6 min

3 marks

A basketball leaves a player's hands $2.05$ m above the floor, travelling at $52°$ above the horizontal. The centre of the hoop is $4.20$ m away horizontally and $3.05$ m above the floor.

Calculate the launch speed needed for the ball to pass through the centre of the hoop. Take $g = 9.8$ m s⁻² and neglect air resistance.

Your answerm s⁻¹

Give it to 3 significant figures.

Hint 1You are given a horizontal distance and a vertical distance but no time, and you want a speed. That shape of problem is solved by eliminating time between the two component equations.

Hint 2Write $x = u cos θ t$ and $Δ y = u sin θ t - \frac{1}{2} g t^{2}$ , then substitute $t = x / (u cos θ)$ into the second.

Hint 3The vertical displacement is $3.05 - 2.05 = 1.00$ m, not $3.05$ m. Everything is measured from the release point, not from the floor.

Where students lose marks on this one

Using $\Delta y=3.05$ m.
Why it happens: The hoop height is quoted from the floor, and the release height is easy to read as scene setting rather than as data.
The equations of motion track displacement from where the object started. Every height in a projectile answer is measured from the launch point, so subtract the release height first.
Solving for time first from the horizontal equation.
Why it happens: Finding $t$ is the habit built up on cliff and ground launch questions, where the vertical direction hands the time over on its own.
Here $t$ and $u$ are both unknown and the horizontal equation contains both, so it cannot be solved alone. Eliminating $t$ between the two equations is the move, and it is the same move behind the range formula.

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The practical investigation

The syllabus asks you to collect primary data in order to validate the relationships. Validating is not the same as illustrating. A demonstration that looks roughly right validates nothing; a test has to be set up so the data could have disagreed and did not.

Worked example5 marks

Measuring the launch speed from the landing points

A ball is released from the same point on a ramp each time so that it leaves the edge of a horizontal bench at the same speed. The bench top is $0.755$ m above the floor. Five trials give landing distances from the bench edge of $0.812$ , $0.798$ , $0.825$ , $0.805$ and $0.810$ m.

Use this primary data to determine the launch speed, and state how the result could be checked independently.

Average the repeated trials and state the spread

A single trial cannot show its own uncertainty. Repeating is what turns five numbers into one number plus a statement of how much to trust it.

$\overset{x}{ˉ} = \frac{0.812 + 0.798 + 0.825 + 0.805 + 0.810}{5} = 0.810 m$

Half the range of the readings gives an estimate of the uncertainty:

$\frac{0.825 - 0.798}{2} = 0.014 m$

So $x = 0.810 \pm 0.014$ m, an uncertainty of $1.7$ per cent.

Find the fall time from the bench height

The ball leaves horizontally, so the vertical column knows nothing about the launch speed and can be solved on its own.

$h = \frac{1}{2} g t^{2} \Rightarrow t = \frac{2 h}{g} = \frac{2 ( 0.755 )}{9.8} = 0.393 s$

Combine them to get the launch speed

The horizontal motion is uniform, so the launch speed is simply the mean distance divided by the fall time.

$u_{x} = \frac{x ˉ}{t} = \frac{0.810}{0.393} = 2.06 m s^{- 1}$

The percentage uncertainty carries through the division, so

$u_{x} = 2.06 \pm 0.04 m s^{- 1}$

Check it against something the model was not used to obtain

Using the projectile model to find a number and then claiming the number validates the model is circular. The test only has force if the same quantity is measured a second way.

Film the ball crossing a $0.500$ m marked section of the bench before the edge. If it takes $0.242$ s, the launch speed is $0.500/0.242 = 2.07$ m s⁻¹.

Two independent routes giving $2.06$ and $2.07$ m s⁻¹, with an uncertainty of $0.04$ , agree. That agreement is the validation, because the two measurements had every opportunity to disagree.

Answer

The launch speed is $2.06 \pm 0.04$ m s⁻¹ from the projectile data, and $2.07$ m s⁻¹ from direct timing on the bench. The agreement within uncertainty supports the model of independent horizontal and vertical motion.

Is that answer sensible?

A ball rolling off a bench at about two metres per second is a gentle roll, which matches a short ramp. And the two methods agreeing to better than one per cent is the right kind of result: close, but not suspiciously identical, which is what real repeated measurement looks like.

What if my data does not agree?

Where to look
First ask whether the disagreement is larger than your uncertainty. A one per cent gap with a three per cent uncertainty is agreement, not a problem.
The idea you need
If it is a real gap, look at its direction. A projectile range that is consistently short of prediction points at friction on the ramp or at air resistance, both of which remove energy. A range consistently long points at a measurement offset, such as measuring from the bench face rather than from below the ball.
How to set it up
Then ask whether the flaw shifts every point the same way or scatters them. Systematic errors preserve the shape of the relationship and move the whole line, which is why an intercept that should be at the origin and is not is the most informative thing on the graph.

Exam question

Harder · about 9 min

5 marks

A student investigates projectile motion by releasing a ball from the same point on a ramp each time, so that it leaves the edge of a horizontal bench at the same speed on every trial. They vary the height $h$ of the bench above the floor and measure the horizontal distance $x$ from the bench edge to the landing point.

Their results were:

$h = 0.40$ m, $x = 0.71$ m
$h = 0.60$ m, $x = 0.87$ m
$h = 0.80$ m, $x = 1.02$ m
$h = 1.00$ m, $x = 1.12$ m
$h = 1.20$ m, $x = 1.24$ m

Explain how this data should be processed and graphed to test the projectile model, what a successful test would look like, and identify one source of systematic error together with its effect on the graph.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Derive the relationship the model predicts between $x$ and $h$ before you decide what to plot.

Hint 2A launch from rest vertically gives $h = \frac{1}{2} g t^{2}$ and $x = u_{x} t$ . Eliminate $t$ .

Hint 3The prediction is $x = u_{x} 2 h / g$ , so $x$ is proportional to $h$ . A proportionality is tested by a straight line through the origin, which means plotting $x$ against $h$ rather than against $h$ .

Where students lose marks on this one

Plotting $x$ against $h$ and reporting that the result is a curve, so the model is confirmed.
Why it happens: The data is plotted in the form it was collected in, and any smooth curve is taken as agreement.
A curve is not evidence for a particular relationship, because many relationships produce curves. Linearise, then the test becomes a specific one the data can fail.
Listing random errors such as parallax when asked for a systematic error.
Why it happens: Both are labelled *error*, and parallax is the most rehearsed example in the course.
A random error scatters points about the line and is reduced by repetition. A systematic error shifts every point the same way and survives any amount of repetition. Only the second can be diagnosed from an intercept.

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Writing about the assumptions

Assesstypically 5 to 8 marks

Demands: Make a judgement of value, size or significance, and support it.
Shape: Criteria, evidence against them, then a stated judgement. The judgement must be visible.
Loses marks: Presenting balanced evidence and never committing to a verdict.

An assess question on air resistance is common enough to prepare for specifically, and it is answered badly for a predictable reason: students describe the effects and never deliver a judgement.

Through a marker’s eyes

4 marks

Assess the validity of the assumption that air resistance is negligible when modelling the flight of a thrown cricket ball. (4 marks)

The attempt

2 out of 4

Air resistance would slow the ball down as it travels. This means the ball would not go as far as the calculation predicts. The path would not be a perfect parabola because the ball is being slowed. In real life air resistance is always present so the model is not completely accurate.

What the marker sees

The response earns the mark for identifying that the range is overestimated, and a second for recognising that the path departs from a parabola. After that it stops.

Three things are missing, and each of them is a mark. There is no mechanism: slows the ball down does not say that the drag acts opposite to the velocity and therefore has a component in both directions, which is what actually breaks the independence the model relies on. There is no sense of size: not completely accurate covers everything from a one per cent error to a factor of three, so it commits to nothing a marker can credit. And there is no judgement. The question says assess, which asks for a verdict on validity, and the closing sentence is a hedge rather than a position.

The final sentence is the most common ending to an assess response and it earns nothing. Every model is imperfect. The question is whether this one is good enough for this purpose.

The same answer, fixed

4 out of 4

Air resistance acts opposite to the ball's velocity at every instant, so it has both a horizontal and a vertical component, and its magnitude grows with speed. This removes the two conditions the model depends on: the horizontal velocity would no longer be constant, and the vertical acceleration would no longer be a steady $9.8$ m s⁻². The path would therefore be an asymmetric curve, steeper on descent than on ascent, landing more slowly and at a steeper angle than it was launched.
For a cricket ball the effect is measurable but modest. The ball is dense and compact and a typical throw is around $25$ m s⁻¹, so the true range falls short of the ideal prediction by roughly five to ten per cent rather than by a factor.
The assumption is therefore valid for the purpose of HSC calculations, which are concerned with the structure of the motion and are quoted to two significant figures. It would not be valid for predicting an actual throw in a match to the nearest metre, and it would be clearly invalid for a lighter or faster projectile such as a shuttlecock, where drag dominates and the path is not parabolic at all.

Exam question

Hardest asked · about 11 min

6 marks

The projectile model used throughout this module assumes zero air resistance.

Assess the effect of this assumption on the predictions the model makes, referring to the shape of the path, the range, and the conditions under which the assumption is reasonable. (6 marks)

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Assess asks for a judgement supported by reasoning, not a list of effects. Decide early what your verdict is and build towards it.

Hint 2Air resistance acts opposite to the velocity, so it has a component in both directions, and its size grows with speed. Work out what each of those two facts does to the model separately.

Hint 3Structure it as three moves: what the assumption removes, what changes as a result, and when the removal is small enough not to matter. A worked comparison with real numbers is what separates a strong response from a correct one.

Where students lose marks on this one

Writing that air resistance slows the projectile down, and stopping there.
Why it happens: *Assess* is read as *describe*, so the response reports an effect and never reaches a judgement.
The mark scheme for an assess question reserves marks for the verdict and its justification. State whether the assumption is acceptable, for what, and why.
Claiming that air resistance affects only the horizontal motion.
Why it happens: It is remembered as the thing that stops the horizontal velocity from being constant, which is the one place the ideal model is obviously fragile.
Drag opposes the velocity, whichever way that points. On the way up it adds to gravity and on the way down it opposes gravity, which is why the fall takes longer than the rise.
Asserting that the path becomes a straight line or a circle.
Why it happens: The parabola is known to be wrong, so it is replaced with another familiar shape rather than described as a distorted parabola.
The path stays curved and continuous. What changes is the symmetry: it steepens through the descent, approaching a vertical drop in the limit where the object reaches terminal velocity.

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The path itself, as one equation

So far the position has always been described by two equations sharing a time. Eliminating the time between them gives the shape of the path directly, with no $t$ in it at all.

Eliminating time to get the trajectory equation

4 steps

1
Start from the two component equations
These are the same two equations used throughout the lesson. Nothing new is being assumed here.
$x = u cos θ t y = u sin θ t - \frac{1}{2} g t^{2}$
2
Solve the horizontal one for time
The horizontal equation is the simpler of the two and contains t linearly, so it is the one to rearrange.
$t = \frac{x}{u c o s θ}$
3
Substitute into the vertical one
Every appearance of t is now replaced by an expression in x, which is what removes time from the description.
$y = u sin θ (\frac{x}{u c o s θ}) - \frac{1}{2} g (\frac{x}{u c o s θ})^{2}$
4
Simplify
The first term collapses to a tangent, which is what makes the shape recognisable.
$y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$
This is a quadratic in $x$ with a negative coefficient on $x^{2}$ , which is the formal statement that the path is a downward parabola. It also explains why the parabola is steeper for a slower launch: $u^{2}$ sits in the denominator of the term that bends the path.

This is the equation to reach for whenever a question names a point the projectile must pass through. It relates $x$ , $y$ , $u$ and $θ$ with no time in sight, so any one of them can be found from the other three. It is not on the formulae sheet, but it can be produced in four lines, and producing it is itself worth marks in a derivation question.

Path of the projectile

Change the launch angle and watch two things at once: where the peak sits, and where the ball lands. They do not move together.

The numbers behind this graph

Time (s)	Horizontal distance (m)	Height (m)
0.00	0.0	0.0
0.72	12.8	10.2
1.44	25.5	15.3
2.16	38.3	15.3
2.89	51.0	10.2
3.61	63.8	0.0

Launch speed25 m s⁻¹

Launch angle45 °

Launch height0 m

Time of flight: 3.61 s
Range: 63.8 m
Greatest height: 15.9 m
Speed on landing: 25.0 m s⁻¹
Angle on landing: 45° below horizontal

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Why 45° stops being the best angle

The greatest range on level ground comes at $45°$ , and the derivation showed why: the range depends on $sin 2 θ$ , which peaks when $2 θ = 90°$ .

Raise the launch point and that stops being true. Set the launch height to $h$ in the trajectory equation, put $y = - h$ at landing, and the angle that maximises $x$ works out to

$sin θ = \frac{1}{2 + \frac{2 g h}{u ^{2}}}$

When $h = 0$ this gives $sin θ = 1/ 2$ , that is, $45°$ , as it must. As $h$ grows the denominator grows, the sine shrinks, and the best angle falls.

The physical account is a trade. Launching steeply buys flight time and spends horizontal speed. Launching flat does the reverse. At $45°$ on level ground the trade is balanced. A launch height hands you extra flight time for nothing, without spending any launch speed on it, so the balance point moves towards keeping more speed horizontal.

The effect is not small at athletic scales. For a shot put released at $2$ m with a launch speed near $13$ m s⁻¹, the optimum is about $42°$ . Elite throwers use angles lower again, around $35°$ , for a reason outside this model entirely: for a human body the achievable release speed itself drops as the angle rises, and release speed matters more than angle because the range depends on its square.

CheckpointAnswer before reading on.

A shot putter releases the shot from $2.0$ m above the ground, and it lands on the ground. Air resistance is negligible.

The launch angle that gives the greatest horizontal distance is:

Aless than $45°$
Bexactly $45°$
Cgreater than $45°$
Dexactly $45°$ , because the range equation contains no height term

Hint 1Check whether the standard range result applies to this situation before using it.

Hint 2 $R = u^{2} sin 2 θ / g$ was derived by setting the vertical displacement to zero. Here it is $- 2.0$ m.

Hint 3The release height already buys extra flight time. Ask whether it is still worth paying for more time by throwing steeply.

Where this model sits in physics

The independence of the two directions is not a special feature of projectiles. It is the vector nature of Newton's second law: $F = m a$ holds component by component, so a force with no horizontal part produces no horizontal acceleration regardless of what else is happening. The same argument is why a charged particle entering a uniform electric field sideways traces a parabola in Module 6, and the algebra there is identical with $q E / m$ in place of $g$ .

What does not carry over is the constant acceleration. Circular motion, the next topic in this module, has an acceleration that changes direction continuously, so the equations of motion used here do not apply to it at all. Gravitational fields, later in the module, have an acceleration that changes with distance, which is why $s = u t + \frac{1}{2} a t^{2}$ gives way to energy methods there.

A projectile is the last situation in the course where the three equations of motion can simply be written down. It is worth noticing that, because the habit of reaching for them survives long past the point where they are valid.