B2.8 Solve projectile-motion problems using horizontal and vertical components

Ontario Grade 11 Physics

Kinematics

Separate horizontal and vertical motion to find where and when a projectile moves.

A projectile is an object that moves through the air after it has been launched. For this lesson, assume air resistance is small enough to ignore. After launch, gravity changes the object’s vertical motion, while its horizontal velocity stays constant. Treat the horizontal and vertical motions separately, but use the same time for both. This method helps you find a projectile’s position, flight time, or velocity without mixing the two directions.

What you will learn

Describe a projectile’s motion using horizontal and vertical components.
Use the vertical motion to find time, then use the horizontal motion to find distance.
Solve projectile problems with horizontal or angled launches and check units, direction, and reasonableness.

1. Prerequisite bridge: vectors and components

A scalar has magnitude only. Time and distance are scalars. A vector has both magnitude and direction. Velocity and displacement are vectors. For example, a velocity can be described as a speed of

10.0\ \mathrm{m/s}

to the right.

A vector can be split into horizontal and vertical components. Components are the parts of a vector along chosen directions. For a launch at angle

\theta

above the horizontal, use right-angle trigonometry: the horizontal component is the launch speed times

\cos\theta

, and the vertical component is the launch speed times

\sin\theta

. The angle must be measured from the horizontal for these relationships.

v_{ix}=v_i\cos\theta, v_{iy}=v_i\sin\theta

Use metres (m), seconds (s), and metres per second (m/s).
A component includes a direction or sign, not just a size.
The horizontal and vertical components describe the same launch velocity.

2. Set up the physical situation and directions

Define the system as the projectile, such as a ball or a package in flight. Use a ground-based reference frame: positions are measured relative to the ground. Choose right as the positive horizontal direction and up as the positive vertical direction. With this choice, gravity points down, so vertical acceleration is negative.

A motion diagram shows the direction of motion and the chosen axes. The arrows for velocity can change direction, but the horizontal arrow stays the same length in this model. The vertical part changes because gravity acts downward.

Ignore air resistance. The only acceleration considered during flight is the acceleration due to gravity,

g=9.8\ \mathrm{m/s^2}

downward. This model is appropriate for solving the projectile-motion problems in this lesson. If a projectile lands below its launch point, its final vertical displacement is negative under the chosen sign convention.

a_x=0, a_y=-9.8\ \mathrm{m/s^2}

The horizontal and vertical components share one elapsed time, $t$ .
Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is $a_y=-9.8\ \mathrm{m/s^2}$ .

3. Use the component equations

Displacement is the change in position. In the horizontal direction, constant velocity means horizontal displacement equals horizontal velocity multiplied by time. In the vertical direction, use the constant-acceleration equation. The symbols

x

and

y

represent displacements from the launch point;

v_{ix}

and

v_{iy}

are the initial velocity components;

v_{yf}

is the final vertical velocity.

A useful strategy is to solve the vertical part first when time is unknown. Vertical displacement and vertical velocity are affected by gravity, so they can reveal the flight time. Then use that same time in the horizontal equation to find horizontal displacement. Keep the signs of vertical quantities consistent with up as positive.

Write down the known values and the unknown before substituting. If a problem gives a launch angle and speed, find both initial components first. If it gives a horizontal launch, the initial vertical velocity is zero. Keep units in substitutions, round the final result to a sensible number of significant figures, and state its direction.

x=v_{ix}t, y=v_{iy}t+

\frac{1}{2}

a_yt^2, v_{yf}=v_{iy}+a_yt

Use the vertical equation to determine time when needed.
Use the same time in both directions.
A negative vertical result means downward or below the starting position.

4. A reliable problem-solving routine

First, define the projectile and reference frame. State the positive directions. Next, list the known values and the unknown. Draw a simple motion sketch with horizontal and vertical axes and label any launch angle, height, or displacement that matters.

Resolve the launch velocity into components if needed. Choose an equation that includes the unknown and known values in one direction. Find the time from vertical information when possible. Use that time horizontally, then report the result with units and direction.

Finally, check the answer. A time should be positive. A horizontal distance should point in the direction of horizontal motion. A downward vertical displacement should be negative with this sign convention. Compare the result with the situation: a longer time at the same horizontal speed should give a greater horizontal distance.

Do not combine horizontal and vertical values in one equation.
Do not confuse speed, which is scalar, with velocity, which has direction.
Check signs, SI units, significant figures, and whether the result makes sense.

Component map for a projectile

Direction	Positive choice	Acceleration	Useful relationship
Horizontal	Right	$0\ \mathrm{m/s^2}$	$x=v_{ix}t$
Vertical	Up	$-9.8\ \mathrm{m/s^2}$	$y=v_{iy}t+\frac{1}{2}a_yt^2$

Worked example

1. Horizontal launch from a ledge

A ball rolls horizontally off a ledge that is

20.0\ \mathrm{m}

high at

12.0\ \mathrm{m/s}

. Find its time in the air and horizontal distance from the ledge when it lands. Ignore air resistance.

Set directions and list values
The system is the ball, viewed from the ground. Choose right and up as positive. The ball launches horizontally, so $v_{ix}=+12.0\ \mathrm{m/s}$ and $v_{iy}=0$ . The landing point is $20.0\ \mathrm{m}$ below launch, so $y=-20.0\ \mathrm{m}$ . The unknowns are flight time and horizontal displacement.
Find the flight time vertically
Use vertical displacement because it includes the known height, the initial vertical velocity, and time. The time must be positive.
$-20.0\ \mathrm{m}=(0\ \mathrm{m/s})t+\frac{1}{2}(-9.8\ \mathrm{m/s^2})t^2$
Use the positive time horizontally
With constant horizontal velocity, the ball covers its horizontal displacement at $12.0\ \mathrm{m/s}$ during the same flight time.
$t=2.02\ \mathrm{s},\quad x=(12.0\ \mathrm{m/s})(2.02\ \mathrm{s})=24.2\ \mathrm{m}$

Answer: The ball is in the air for

2.02\ \mathrm{s}

and lands

24.2\ \mathrm{m}

horizontally from the ledge, to the right.

Check: The time is positive. The horizontal units reduce to metres. A fall of

20.0\ \mathrm{m}

taking about

2\ \mathrm{s}

is reasonable, and a speed of

12.0\ \mathrm{m/s}

over that time gives a distance near

24\ \mathrm{m}

.

Worked example

2. Launch and landing at the same height

A ball is launched at

18.0\ \mathrm{m/s}

at

35.0^\circ

above the horizontal. It lands at its launch height. Find its flight time and horizontal range. Ignore air resistance.

Resolve the launch velocity
The system is the ball in a ground-based frame. Choose right and up as positive. The launch speed is a scalar; its components are velocity vectors along the chosen axes.
$v_{ix}=(18.0\ \mathrm{m/s})\cos35.0^\circ=14.7\ \mathrm{m/s},\quad v_{iy}=(18.0\ \mathrm{m/s})\sin35.0^\circ=10.3\ \mathrm{m/s}$
Find the flight time vertically
The projectile returns to its launch height, so its vertical displacement is zero. Use the vertical displacement equation and take the nonzero time; the zero-time solution is the launch instant.
$0=(10.3\ \mathrm{m/s})t+\frac{1}{2}(-9.8\ \mathrm{m/s^2})t^2,\quad t=2.10\ \mathrm{s}$
Find the horizontal range
Horizontal velocity remains constant. Multiply the horizontal component by the flight time to find the horizontal displacement.
$x=(14.7\ \mathrm{m/s})(2.10\ \mathrm{s})=30.9\ \mathrm{m}$

Answer: The flight time is

2.10\ \mathrm{s}

, and the ball lands about

30.9\ \mathrm{m}

to the right of its launch point.

Check: The displacement units are metres. The time is positive and the range is in the launch direction. A horizontal component near

15\ \mathrm{m/s}

over about

2\ \mathrm{s}

gives a range near

30\ \mathrm{m}

.

Worked example

3. Angled launch from a height

A ball is launched from a platform

15.0\ \mathrm{m}

above the ground at

14.0\ \mathrm{m/s}

and

30.0^\circ

above the horizontal. Find its time to reach the ground and its horizontal distance from the platform’s edge. Ignore air resistance.

Set directions and components
The system is the ball, viewed from the ground. Choose right and up as positive. The ground is $15.0\ \mathrm{m}$ below the launch point, so $y=-15.0\ \mathrm{m}$ . Resolve the initial velocity into components.
$v_{ix}=(14.0\ \mathrm{m/s})\cos30.0^\circ=12.1\ \mathrm{m/s},\quad v_{iy}=(14.0\ \mathrm{m/s})\sin30.0^\circ=7.00\ \mathrm{m/s}$
Solve for time using vertical motion
Substitute the vertical displacement, initial vertical velocity, and gravitational acceleration. The positive solution is the time when the ball reaches the ground; reject a negative time because it would describe a time before launch.
$-15.0\ \mathrm{m}=(7.00\ \mathrm{m/s})t+\frac{1}{2}(-9.8\ \mathrm{m/s^2})t^2,\quad t=2.60\ \mathrm{s}$
Find horizontal displacement
Use the same flight time with the constant horizontal component. Report the distance in the direction of launch.
$x=(12.1\ \mathrm{m/s})(2.60\ \mathrm{s})=31.5\ \mathrm{m}$

Answer: The ball reaches the ground after

2.60\ \mathrm{s}

and lands

31.5\ \mathrm{m}

to the right of the platform’s edge.

Check: The time is positive, and the horizontal distance has units of metres. The ball rises at first but lands below its launch point, so a flight lasting more than

2\ \mathrm{s}

is reasonable.

Common mistakes and how to avoid them

Using the full launch speed as the horizontal velocity for an angled launch.

Correction: Resolve the launch velocity first. The horizontal component is

v_i\cos\theta

when the angle is measured from the horizontal.

Giving gravity a positive sign while up is positive.

Correction: With up chosen as positive, gravity points down, so use

a_y=-9.8\ \mathrm{m/s^2}

.

Using a different time in the horizontal and vertical calculations.

Correction: Both components describe the same projectile during the same flight, so they use the same elapsed time.

Calling a negative vertical displacement an error.

Correction: A negative value means the final position is below the launch point when up is positive.

Lesson summary

Describe a launch velocity using horizontal and vertical components.
Use vertical motion to find time when the vertical information is known.
Use that same time with constant horizontal velocity to find horizontal displacement.
Include directions and SI units, then check the sign, significant figures, and physical reasonableness.

Check your understanding

Question 1

A projectile is launched horizontally to the right. Ignoring air resistance, which statement describes its horizontal motion?

Its horizontal velocity stays constant to the right.
Its horizontal velocity increases downward.
Its horizontal velocity becomes zero as soon as it leaves the launcher.
correctIndex

Show answer and explanation

Its horizontal velocity stays constant to the right.

Horizontal acceleration is zero in this model, so the horizontal velocity remains constant. Gravity changes the vertical component.

Question 2

A ball is launched from a point and lands

8.0\ \mathrm{m}

below it. With up chosen as positive, what is its vertical displacement?

$+8.0\ \mathrm{m}$
$-8.0\ \mathrm{m}$
$0\ \mathrm{m}$
correctIndex

Show answer and explanation

-8.0\ \mathrm{m}

Down is negative when up is positive, so the vertical displacement is

-8.0\ \mathrm{m}

.

Question 3

A projectile’s horizontal component is

6.0\ \mathrm{m/s}

and its flight time is

3.0\ \mathrm{s}

. What is its horizontal displacement?

$2.0\ \mathrm{m}$ to the right
$18\ \mathrm{m}$ to the right
$18\ \mathrm{m}$ downward
correctIndex

Show answer and explanation

18\ \mathrm{m}

to the right

Horizontal displacement is horizontal velocity multiplied by time:

6.0\ \mathrm{m/s}

times

3.0\ \mathrm{s}

gives

18\ \mathrm{m}

to the right.

Key terms

Projectile: An object moving through the air after launch, treated here as affected only by gravity.
Component: The part of a vector along one chosen direction, such as horizontal or vertical.
Displacement: The change in position, including direction.
Reference frame: The point of view used to describe an object's position and motion.
Range: The horizontal displacement from launch to landing.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.8. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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B2.8 · Solve projectile-motion problems using horizontal and vertical components

Ontario Grade 11 Physics

Separate horizontal and vertical motion to find where and when a projectile moves.

What you will learn

1. Prerequisite bridge: vectors and components

2. Set up the physical situation and directions

3. Use the component equations

4. A reliable problem-solving routine

Component map for a projectile

1. Horizontal launch from a ledge

2. Launch and landing at the same height

3. Angled launch from a height

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review