Projectile motion

What you'll learn

Why a projectile’s horizontal and vertical motions can be treated independently.
How to resolve an initial velocity into perpendicular components.
How to use constant-acceleration equations separately in each direction.
How to solve flight time, range, maximum height and impact velocity problems.

The starting point: motion in two dimensions

In earlier motion work, you often used one straight line: left/right or up/down. Projectile motion needs two perpendicular directions at once.

A displacement is the change in position of an object, with both size and direction. A velocity is the rate of change of displacement. An acceleration is the rate of change of velocity.

For projectile motion, we usually choose:

the horizontal direction as the $x$ -direction
the vertical direction as the $y$ -direction

These directions are perpendicular, meaning they are at right angles to each other.

Definition

Projectile

A projectile is an object that has been launched and then moves only under gravity. In this A-Level model, air resistance is ignored unless a question explicitly tells you otherwise.

Resolving the initial velocity

A component of a vector is the part of that vector in a chosen direction. Resolving a vector means splitting it into components, usually horizontal and vertical.

If a projectile is launched with speed $u$ at an angle $θ\theta$ above the horizontal, then:

ux=ucos⁡θuy=usin⁡θ\begin{aligned} u_x &= u\cos\theta \\ u_y &= u\sin\theta \end{aligned}

The horizontal component uses cosine because it is adjacent to the angle. The vertical component uses sine because it is opposite the angle.

Example

Resolving launch velocity

A ball is launched at $\ \text{m s}^{-1}$ at $35.0∘35.0^\circ$ above the horizontal. Find its initial horizontal and vertical velocity components.

Choose horizontal and vertical axes. The launch angle is measured from the horizontal, so the horizontal component is adjacent to the angle and the vertical component is opposite it.
Calculate the horizontal component: $s−1u_x = u\cos\theta = (18.0 \ \text{m s}^{-1})\cos 35.0^\circ = 14.7 \ \text{m s}^{-1}$ .
Calculate the vertical component: $s−1u_y = u\sin\theta = (18.0 \ \text{m s}^{-1})\sin 35.0^\circ = 10.3 \ \text{m s}^{-1}$ .

The key model: horizontal and vertical motion are independent

Once the projectile is in flight, gravity acts vertically downwards. In the no-air-resistance model, there is no horizontal acceleration.

So the projectile has:

constant velocity horizontally
constant acceleration vertically

Near Earth’s surface, the acceleration due to gravity has magnitude $\ \text{m s}^{-2}$ . If you choose upward as positive, then the vertical acceleration is $s−2a_y = -g = -9.81 \ \text{m s}^{-2}$ .

Projectile path showing resolved components, constant horizontal velocity and downward acceleration

Key Idea

Independence of components

The horizontal and vertical motions share the same time $t$ , but the equations are applied separately in each direction. Vertical acceleration does not change the horizontal velocity.

For horizontal motion:

ax=0vx=uxsx=uxt\begin{aligned} a_x &= 0 \\ v_x &= u_x \\ s_x &= u_x t \end{aligned}

For vertical motion, use the constant-acceleration equations with $a_y = -g$ if upward is positive:

v=u+ats=ut+12at2v2=u2+2ass=(u+v2)t\begin{aligned} v &= u + at \\ s &= ut + \frac{1}{2}at^2 \\ v^2 &= u^2 + 2as \\ s &= \left(\frac{u+v}{2}\right)t \end{aligned}

When using these vertically, replace $u$ , $v$ , $a$ and $s$ with $u_y$ , $v_y$ , $a_y$ and $s_y$ .

Example

Horizontal launch from a height

A ball rolls horizontally off a table at $\ \text{m s}^{-1}$ . The table is $\ \text{m}$ high. Find the time taken to hit the floor and the horizontal distance travelled.

Choose upward as positive. Vertically, the ball starts with $s−1u_y = 0 \ \text{m s}^{-1}$ , has displacement $ms_y = -1.25 \ \text{m}$ , and acceleration $s−2a_y = -9.81 \ \text{m s}^{-2}$ .
Use the vertical equation $sy=uyt+12ayt2s_y = u_y t + \frac{1}{2}a_y t^2$ : $\ \text{m} = 0 + \frac{1}{2}(-9.81 \ \text{m s}^{-2})t^2$ , giving $\ \text{s}$ .
Use the same time horizontally. Since $a_x = 0$ , $ms_x = u_x t = (6.0 \ \text{m s}^{-1})(0.505 \ \text{s}) = 3.03 \ \text{m}$ , so the horizontal distance is about $\ \text{m}$ .

Common Mistake

Mixing components

Do not put horizontal and vertical quantities into the same SUVAT equation. For example, never combine horizontal displacement with vertical acceleration in one equation.

What the graphs look like

Because the horizontal velocity is constant, the horizontal displacement increases linearly with time. Because the vertical acceleration is constant, the vertical velocity changes linearly with time, and the vertical displacement forms a curved graph.

Projectile motion graphs for horizontal displacement, vertical displacement, horizontal velocity and vertical velocity

At the highest point of the flight, the vertical velocity is zero. This does not mean the projectile has stopped.

Common Mistake

Top of the path

At the highest point, only $v_y$ is zero. The horizontal velocity $v_x$ is still unchanged, and the acceleration is still $g$ downward.

Example

Using a vertical velocity graph

A projectile has vertical velocity $\ \text{m s}^{-1}$ at launch. Its vertical velocity falls to zero after $\ \text{s}$ . Find the vertical acceleration and the maximum height gained.

Use the gradient of the vertical velocity-time graph: $s−2a_y = \frac{\Delta v_y}{\Delta t} = \frac{0 - 14.7 \ \text{m s}^{-1}}{1.50 \ \text{s}} = -9.80 \ \text{m s}^{-2}$ .
The vertical displacement is the area under the vertical velocity-time graph. The area is triangular, so $ms_y = \frac{1}{2}(1.50 \ \text{s})(14.7 \ \text{m s}^{-1}) = 11.0 \ \text{m}$ .
Since $v_y = 0$ at this instant, this displacement is the maximum height above the launch point.

Solving a full projectile problem

A good method is:

Draw a simple sketch and choose positive directions.
Resolve the initial velocity into $u_x$ and $u_y$ .
Use vertical motion to find the time, because the height information is usually vertical.
Use that same time in the horizontal equation.
If asked for final velocity, combine $v_x$ and $v_y$ using Pythagoras and trigonometry.

The time of flight is the total time the projectile is in the air. The range is the horizontal distance travelled before landing.

Tip

Best order

Most projectile problems become easier if you find the time from the vertical motion first, then use that time horizontally.

Example

Range and impact velocity for an angled launch

A ball is launched from ground level at $\ \text{m s}^{-1}$ at $30.0∘30.0^\circ$ above the horizontal. It lands back at ground level. Find the time of flight, range, maximum height and impact velocity.

Resolve the initial velocity: $s−1u_x = (20.0 \ \text{m s}^{-1})\cos 30.0^\circ = 17.3 \ \text{m s}^{-1}$ and $s−1u_y = (20.0 \ \text{m s}^{-1})\sin 30.0^\circ = 10.0 \ \text{m s}^{-1}$ .
Use vertical motion to find the time of flight. Since it lands at the same height, $ms_y = 0 \ \text{m}$ , so $\ \text{m s}^{-1})t + \frac{1}{2}(-9.81 \ \text{m s}^{-2})t^2$ . The non-zero solution is $\ \text{s}$ .
Use horizontal motion to find the range: $ms_x = u_x t = (17.3 \ \text{m s}^{-1})(2.04 \ \text{s}) = 35.3 \ \text{m}$ .
Find the maximum height using $v_y^2 = u_y^2 + 2a_y s_y$ , with $s−1v_y = 0 \ \text{m s}^{-1}$ at the top: $\ \text{m s}^{-1})^2 + 2(-9.81 \ \text{m s}^{-2})s_y$ , so $ms_y = 5.10 \ \text{m}$ .
Find the impact velocity components. Horizontally, $s−1v_x = 17.3 \ \text{m s}^{-1}$ . Vertically, $s−1v_y = u_y + a_y t = 10.0 \ \text{m s}^{-1} - (9.81 \ \text{m s}^{-2})(2.04 \ \text{s}) = -10.0 \ \text{m s}^{-1}$ . The speed is $\sqrt{(17.3 \ \text{m s}^{-1})^2 + (-10.0 \ \text{m s}^{-1})^2} = 20.0 \ \text{m s}^{-1}$ , directed $30.0∘30.0^\circ$ below the horizontal.

Common Mistake

When the model breaks

If air resistance is included, the horizontal velocity is no longer constant and these simple projectile equations no longer apply directly.

Exam technique

In the exam

Draw axes and mark $a_x = 0$ and $a_y = -g$ if upward is positive.
Keep horizontal and vertical quantities separate, but remember they share the same time $t$ .
Do not assume the projectile lands at the launch height unless the question says or shows it.
Quote velocities with direction or components, not just a speed, when the question asks for velocity.

Self review

Check yourself

At the highest point of a projectile’s path, which velocity component is zero and which is not?
Why can the time found from the vertical motion be used in the horizontal motion?
A ball is launched horizontally from a cliff. Which equation would you use first to find its time of flight?

What you'll learn

Why a projectile’s horizontal and vertical motions can be treated independently.
How to resolve an initial velocity into perpendicular components.
How to use constant-acceleration equations separately in each direction.
How to solve flight time, range, maximum height and impact velocity problems.

The starting point: motion in two dimensions

In earlier motion work, you often used one straight line: left/right or up/down. Projectile motion needs two perpendicular directions at once.

A displacement is the change in position of an object, with both size and direction. A velocity is the rate of change of displacement. An acceleration is the rate of change of velocity.

For projectile motion, we usually choose:

the horizontal direction as the $x$ -direction
the vertical direction as the $y$ -direction

These directions are perpendicular, meaning they are at right angles to each other.

Definition

Projectile

A projectile is an object that has been launched and then moves only under gravity. In this A-Level model, air resistance is ignored unless a question explicitly tells you otherwise.

Resolving the initial velocity

A component of a vector is the part of that vector in a chosen direction. Resolving a vector means splitting it into components, usually horizontal and vertical.

If a projectile is launched with speed $u$ at an angle $θ\theta$ above the horizontal, then:

ux=ucos⁡θuy=usin⁡θ\begin{aligned} u_x &= u\cos\theta \\ u_y &= u\sin\theta \end{aligned}

The horizontal component uses cosine because it is adjacent to the angle. The vertical component uses sine because it is opposite the angle.

Example

Resolving launch velocity

A ball is launched at $\ \text{m s}^{-1}$ at $35.0∘35.0^\circ$ above the horizontal. Find its initial horizontal and vertical velocity components.

Choose horizontal and vertical axes. The launch angle is measured from the horizontal, so the horizontal component is adjacent to the angle and the vertical component is opposite it.
Calculate the horizontal component: $s−1u_x = u\cos\theta = (18.0 \ \text{m s}^{-1})\cos 35.0^\circ = 14.7 \ \text{m s}^{-1}$ .
Calculate the vertical component: $s−1u_y = u\sin\theta = (18.0 \ \text{m s}^{-1})\sin 35.0^\circ = 10.3 \ \text{m s}^{-1}$ .

The key model: horizontal and vertical motion are independent

Once the projectile is in flight, gravity acts vertically downwards. In the no-air-resistance model, there is no horizontal acceleration.

So the projectile has:

constant velocity horizontally
constant acceleration vertically

Projectile path showing resolved components, constant horizontal velocity and downward acceleration

Key Idea

Independence of components

The horizontal and vertical motions share the same time $t$ , but the equations are applied separately in each direction. Vertical acceleration does not change the horizontal velocity.

For horizontal motion:

ax=0vx=uxsx=uxt\begin{aligned} a_x &= 0 \\ v_x &= u_x \\ s_x &= u_x t \end{aligned}

For vertical motion, use the constant-acceleration equations with $a_y = -g$ if upward is positive:

v=u+ats=ut+12at2v2=u2+2ass=(u+v2)t\begin{aligned} v &= u + at \\ s &= ut + \frac{1}{2}at^2 \\ v^2 &= u^2 + 2as \\ s &= \left(\frac{u+v}{2}\right)t \end{aligned}

When using these vertically, replace $u$ , $v$ , $a$ and $s$ with $u_y$ , $v_y$ , $a_y$ and $s_y$ .

Example

Horizontal launch from a height

A ball rolls horizontally off a table at $\ \text{m s}^{-1}$ . The table is $\ \text{m}$ high. Find the time taken to hit the floor and the horizontal distance travelled.

Choose upward as positive. Vertically, the ball starts with $s−1u_y = 0 \ \text{m s}^{-1}$ , has displacement $ms_y = -1.25 \ \text{m}$ , and acceleration $s−2a_y = -9.81 \ \text{m s}^{-2}$ .
Use the vertical equation $sy=uyt+12ayt2s_y = u_y t + \frac{1}{2}a_y t^2$ : $\ \text{m} = 0 + \frac{1}{2}(-9.81 \ \text{m s}^{-2})t^2$ , giving $\ \text{s}$ .
Use the same time horizontally. Since $a_x = 0$ , $ms_x = u_x t = (6.0 \ \text{m s}^{-1})(0.505 \ \text{s}) = 3.03 \ \text{m}$ , so the horizontal distance is about $\ \text{m}$ .

Common Mistake

Mixing components

Do not put horizontal and vertical quantities into the same SUVAT equation. For example, never combine horizontal displacement with vertical acceleration in one equation.

What the graphs look like

Projectile motion graphs for horizontal displacement, vertical displacement, horizontal velocity and vertical velocity

At the highest point of the flight, the vertical velocity is zero. This does not mean the projectile has stopped.

Common Mistake

Top of the path

At the highest point, only $v_y$ is zero. The horizontal velocity $v_x$ is still unchanged, and the acceleration is still $g$ downward.

Example

Using a vertical velocity graph

A projectile has vertical velocity $\ \text{m s}^{-1}$ at launch. Its vertical velocity falls to zero after $\ \text{s}$ . Find the vertical acceleration and the maximum height gained.

Use the gradient of the vertical velocity-time graph: $s−2a_y = \frac{\Delta v_y}{\Delta t} = \frac{0 - 14.7 \ \text{m s}^{-1}}{1.50 \ \text{s}} = -9.80 \ \text{m s}^{-2}$ .
The vertical displacement is the area under the vertical velocity-time graph. The area is triangular, so $ms_y = \frac{1}{2}(1.50 \ \text{s})(14.7 \ \text{m s}^{-1}) = 11.0 \ \text{m}$ .
Since $v_y = 0$ at this instant, this displacement is the maximum height above the launch point.

Solving a full projectile problem

A good method is:

Draw a simple sketch and choose positive directions.
Resolve the initial velocity into $u_x$ and $u_y$ .
Use vertical motion to find the time, because the height information is usually vertical.
Use that same time in the horizontal equation.
If asked for final velocity, combine $v_x$ and $v_y$ using Pythagoras and trigonometry.

The time of flight is the total time the projectile is in the air. The range is the horizontal distance travelled before landing.

Tip

Best order

Most projectile problems become easier if you find the time from the vertical motion first, then use that time horizontally.

Example

Range and impact velocity for an angled launch

Resolve the initial velocity: $s−1u_x = (20.0 \ \text{m s}^{-1})\cos 30.0^\circ = 17.3 \ \text{m s}^{-1}$ and $s−1u_y = (20.0 \ \text{m s}^{-1})\sin 30.0^\circ = 10.0 \ \text{m s}^{-1}$ .
Use vertical motion to find the time of flight. Since it lands at the same height, $ms_y = 0 \ \text{m}$ , so $\ \text{m s}^{-1})t + \frac{1}{2}(-9.81 \ \text{m s}^{-2})t^2$ . The non-zero solution is $\ \text{s}$ .
Use horizontal motion to find the range: $ms_x = u_x t = (17.3 \ \text{m s}^{-1})(2.04 \ \text{s}) = 35.3 \ \text{m}$ .
Find the maximum height using $v_y^2 = u_y^2 + 2a_y s_y$ , with $s−1v_y = 0 \ \text{m s}^{-1}$ at the top: $\ \text{m s}^{-1})^2 + 2(-9.81 \ \text{m s}^{-2})s_y$ , so $ms_y = 5.10 \ \text{m}$ .
Find the impact velocity components. Horizontally, $s−1v_x = 17.3 \ \text{m s}^{-1}$ . Vertically, $s−1v_y = u_y + a_y t = 10.0 \ \text{m s}^{-1} - (9.81 \ \text{m s}^{-2})(2.04 \ \text{s}) = -10.0 \ \text{m s}^{-1}$ . The speed is $\sqrt{(17.3 \ \text{m s}^{-1})^2 + (-10.0 \ \text{m s}^{-1})^2} = 20.0 \ \text{m s}^{-1}$ , directed $30.0∘30.0^\circ$ below the horizontal.

Common Mistake

When the model breaks

If air resistance is included, the horizontal velocity is no longer constant and these simple projectile equations no longer apply directly.

Exam technique

In the exam

Draw axes and mark $a_x = 0$ and $a_y = -g$ if upward is positive.
Keep horizontal and vertical quantities separate, but remember they share the same time $t$ .
Do not assume the projectile lands at the launch height unless the question says or shows it.
Quote velocities with direction or components, not just a speed, when the question asks for velocity.

Self review

Check yourself

At the highest point of a projectile’s path, which velocity component is zero and which is not?
Why can the time found from the vertical motion be used in the horizontal motion?
A ball is launched horizontally from a cliff. Which equation would you use first to find its time of flight?

What you'll learn

The starting point: motion in two dimensions

Resolving the initial velocity

The key model: horizontal and vertical motion are independent

What the graphs look like

Solving a full projectile problem

Review Projectile motion by teaching Genie

Remember key concepts with flashcards

Projectile motion

What you'll learn

The starting point: motion in two dimensions

Resolving the initial velocity

The key model: horizontal and vertical motion are independent

What the graphs look like

Solving a full projectile problem

Review Projectile motion by teaching Genie

Remember key concepts with flashcards

Projectile motion

What you'll learn

The starting point: motion in two dimensions

Resolving the initial velocity

The key model: horizontal and vertical motion are independent

What the graphs look like

Solving a full projectile problem

Review Projectile motion by teaching Genie

Remember key concepts with flashcards

Projectile motion Revision Guide

Projectile motion

What you'll learn

The starting point: motion in two dimensions

Resolving the initial velocity

The key model: horizontal and vertical motion are independent

What the graphs look like

Solving a full projectile problem

Review Projectile motion by teaching Genie

Remember key concepts with flashcards

Projectile motion Revision Guide