B2.2 Solve projectile and relative-motion problems with two-dimensional vectors

Ontario Grade 12 Physics

Dynamics

SPH4U B2.2 — separating motion into perpendicular directions

In SPH3U, you used displacement, velocity, and acceleration to describe motion. Displacement and velocity are vectors: each has a magnitude and a direction. Time and speed are scalars: they have magnitude but no direction. This lesson uses these ideas in two dimensions. A projectile follows a curved path, but its horizontal and vertical motions can be solved separately. Relative motion compares an object's motion as seen from different reference frames.

What you will learn

Resolve a vector into horizontal and vertical components.
Solve projectile motion by treating horizontal and vertical motion separately.
Add velocity vectors to describe relative motion between objects and reference frames.
Report answers with suitable units, significant figures, and directions.

1. Set the frame, directions, and vector components

A physical system is the object or objects whose motion you are studying. A reference frame is the viewpoint used to describe their positions and velocities. State the frame before solving. For a projectile near Earth's surface, a useful frame is the ground. For a boat, you may need both the water frame and the shore frame.

Choose coordinate directions and keep them throughout the solution. For projectile motion, choose right as positive

x

and up as positive

y

. Then gravity points in the negative

y

direction. A vector's components are its signed parts along the coordinate axes. Its magnitude is its length, and its direction tells how it is oriented.

If a vector of magnitude

A

makes an angle

\theta

above the positive horizontal direction, its horizontal component is

A_x=A\cos\theta

and its vertical component is

A_y=A\sin\theta

. Include the signs that match your axes. The magnitude can be recovered from perpendicular components using

A=\sqrt{A_x^2+A_y^2}

.

A_x=A\cos\theta,\quad A_y=A\sin\theta

State the physical system and reference frame.
Choose positive directions before assigning signs.
A vector needs both magnitude and direction; a scalar does not.

2. Model projectile motion in two directions

A projectile is an object moving through the air after it has been launched, when air resistance is ignored. After launch, gravity is the only acceleration in this model. Near Earth's surface, its magnitude is about

9.8\,\mathrm{m/s^2}

downward. With up positive, the vertical acceleration is

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

, while horizontal acceleration is zero.

The horizontal and vertical motions share the same elapsed time, but have different accelerations. Horizontally, velocity stays constant. Vertically, gravity changes the velocity. Use the constant-acceleration relationships separately in each direction. The vertical relationship is

\Delta y=v_{iy}t+\frac{1}{2}a_yt^2

, where

\Delta y

is vertical displacement and

v_{iy}

is initial vertical velocity. Horizontally,

\Delta x=v_xt

.

A useful motion sketch labels the launch point, axes, initial velocity components, and gravitational acceleration. For a launch at angle

\theta

, resolve the initial velocity first. For a horizontal launch, the initial vertical velocity is zero. At the top of a curved path, vertical velocity is momentarily zero, but gravity still acts downward.

Choose an equation that contains the unknown and the known quantities. Solve for time from the vertical motion when the vertical displacement is known, then use that time in the horizontal relationship. Keep signs consistent. A negative vertical displacement means the object ends below its starting height.

\Delta x=v_xt,\quad \Delta y=v_{iy}t+\frac{1}{2}a_yt^2

In the ideal projectile model, $a_x=0$ and $a_y=-9.8\,\mathrm{m/s^2}$ for up-positive axes.
Horizontal and vertical calculations use the same time.
Use metres, seconds, metres per second, and metres per second squared.

3. Add velocities for relative motion

Relative velocity describes how fast an object moves as observed in one frame compared with another. Use a clear frame label. For example,

\vec v_{B/G}

means the velocity of boat

B

relative to ground

G

. The velocity of the boat relative to ground equals its velocity relative to water plus the water's velocity relative to ground.

This is vector addition. Draw arrows head-to-tail, or add the horizontal components and vertical components separately. If two components are perpendicular, find the resultant magnitude with the Pythagorean relationship. Then use trigonometry to state its direction. The direction must be described relative to a named axis or compass direction.

Relative-motion questions may also ask where an object arrives. Find the time from motion in the direction of travel across a gap, then use the velocity component along the other direction to find drift. A velocity relative to the water is not automatically the velocity relative to the shore.

\vec v_{B/G}=\vec v_{B/W}+\vec v_{W/G}

Name both frames in relative-velocity notation.
Add velocity components with their signs.
Report resultant velocity with magnitude and direction.

4. Check the result

Before calculating, list known values and the unknown, with SI units. During substitution, carry units through the calculation. Round only the final result, using precision that matches the given values. A direction should agree with the signs of the components: a negative vertical component means downward when up is positive.

Finish with a reasonableness check. A time should not be negative for a forward-moving event. A resultant vector should not be shorter than either of two perpendicular component vectors. In projectile motion, the horizontal distance should equal horizontal speed multiplied by flight time. These checks can reveal a sign, component, or frame error.

v=\sqrt{v_x^2+v_y^2}

Show the relationship, substitution, and final units.
Check signs, direction, and whether the result is physically reasonable.
Round sensibly and do not report more precision than the input supports.

Worked example

Horizontal launch from a ledge

A ball leaves a ledge horizontally at

12.0\,\mathrm{m/s}

. The ledge is

45.0\,\mathrm{m}

above level ground. Find the flight time, horizontal distance, and velocity just before landing. Ignore air resistance.

Define the system and axes
The system is the ball, and the reference frame is the ground. Choose right as positive $x$ and up as positive $y$ . The launch point is the origin, so the landing displacement is $\Delta y=-45.0\,\mathrm{m}$ . The unknowns are flight time and landing velocity.
Find the flight time
The initial vertical velocity is zero because the launch is horizontal. Use vertical displacement to find time, with gravity negative in the chosen frame.
$-45.0\,\mathrm{m}=0\,\mathrm{m/s}\cdot t+\frac{1}{2}(-9.8\,\mathrm{m/s^2})t^2\;\Rightarrow\;t=3.03\,\mathrm{s}$
Find horizontal distance and landing velocity
Horizontal velocity remains $12.0\,\mathrm{m/s}$ . The vertical landing velocity follows from the constant-acceleration relationship. The negative sign indicates downward motion.
$\Delta x=(12.0\,\mathrm{m/s})(3.03\,\mathrm{s})=36.4\,\mathrm{m};\quad v_y=0+(-9.8\,\mathrm{m/s^2})(3.03\,\mathrm{s})=-29.7\,\mathrm{m/s}$
Combine the velocity components
The landing velocity has components $12.0\,\mathrm{m/s}$ right and $29.7\,\mathrm{m/s}$ down. Its magnitude and angle below the horizontal describe the full vector.
$v=\sqrt{(12.0)^2+(29.7)^2}=32.0\,\mathrm{m/s};\quad \theta=\tan^{-1}(29.7/12.0)=68.0^\circ$

Answer: The ball is in flight for

3.03\,\mathrm{s}

and lands

36.4\,\mathrm{m}

horizontally from the ledge. Just before landing, its velocity is

32.0\,\mathrm{m/s}

at

68.0^\circ

below the horizontal, toward the right.

Check: Time is positive. The horizontal distance equals

12.0\,\mathrm{m/s}

times

3.03\,\mathrm{s}

. The downward velocity agrees with the negative vertical direction, and the velocity magnitude is greater than either component.

Worked example

Angled launch and return to the same height

A ball is launched from level ground at

20.0\,\mathrm{m/s}

,

30.0^\circ

above the horizontal. Find its time in the air, horizontal range, and maximum height. Ignore air resistance.

Define the system and resolve launch velocity
The system is the ball in the ground frame. Choose right and up as positive. The initial velocity components come from resolving the launch vector.
$v_{ix}=(20.0\,\mathrm{m/s})\cos30.0^\circ=17.3\,\mathrm{m/s};\quad v_{iy}=(20.0\,\mathrm{m/s})\sin30.0^\circ=10.0\,\mathrm{m/s}$
Find the time to the top and total flight time
At the highest point, vertical velocity is zero. Use the vertical velocity relationship to find the time to the top. Since the ball returns to its launch height in this model, the total flight time is twice this value.
$0=10.0\,\mathrm{m/s}+(-9.8\,\mathrm{m/s^2})t_{\mathrm{top}}\;\Rightarrow\;t_{\mathrm{top}}=1.02\,\mathrm{s};\quad t_{\mathrm{flight}}=2.04\,\mathrm{s}$
Find range and maximum height
Use the total time with constant horizontal velocity for range. Use the time to the top in the vertical displacement relationship for maximum height.
$\Delta x=(17.3\,\mathrm{m/s})(2.04\,\mathrm{s})=35.3\,\mathrm{m};\quad \Delta y_{\mathrm{top}}=(10.0)(1.02)+\frac{1}{2}(-9.8)(1.02)^2=5.10\,\mathrm{m}$

Answer: The ball is in the air for

2.04\,\mathrm{s}

. Its horizontal range is

35.3\,\mathrm{m}

, and its maximum height above the launch point is

5.10\,\mathrm{m}

.

Check: The range has units of metres, and the peak height is positive. The horizontal component stays constant, while the vertical component reaches zero at the top.

Worked example

A boat crossing a river

A boat moves north at

4.0\,\mathrm{m/s}

relative to the water. The water moves east at

3.0\,\mathrm{m/s}

relative to the shore. The river is

120\,\mathrm{m}

wide. Find the boat's velocity relative to shore, its crossing time, and its eastward drift.

Define frames and directions
The system is the boat. Use the shore as the ground frame, east as positive $x$ , and north as positive $y$ . The boat's velocity relative to water is north; the water's velocity relative to shore is east.
$\vec v_{B/W}=(0,4.0)\,\mathrm{m/s};\quad \vec v_{W/G}=(3.0,0)\,\mathrm{m/s}$
Add the velocities
Add corresponding components to obtain the boat's velocity relative to shore. Because the components are perpendicular, use the right-triangle relationships for magnitude and direction.
$\vec v_{B/G}=(3.0,4.0)\,\mathrm{m/s};\quad v=5.0\,\mathrm{m/s};\quad \theta=\tan^{-1}(3.0/4.0)=37^\circ$
Find crossing time and drift
The northward component carries the boat across the river. Divide the width by that component to get time, then use the eastward component for drift.
$t=(120\,\mathrm{m})/(4.0\,\mathrm{m/s})=30\,\mathrm{s};\quad \Delta x=(3.0\,\mathrm{m/s})(30\,\mathrm{s})=90\,\mathrm{m}$

Answer: The boat moves at

5.0\,\mathrm{m/s}

,

37^\circ

east of north relative to shore. It takes

30\,\mathrm{s}

to cross and drifts

90\,\mathrm{m}

east.

Check: The shore-frame velocity includes both water motion and boat motion through the water. The resultant speed exceeds either perpendicular component, and the drift is east, matching the water's motion.

Common mistakes and how to avoid them

Using the full launch speed as the horizontal or vertical velocity.

Correction: Resolve the launch velocity into components before using motion relationships.

Using positive gravity when up is positive.

Correction: Use

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

when the positive vertical direction is up.

Assuming horizontal and vertical motions have separate elapsed times.

Correction: Use one shared time for both components of the same motion.

Treating velocity relative to water as velocity relative to shore.

Correction: Add the water-relative-to-shore velocity as a vector, with frame labels.

Giving a vector answer without a direction.

Correction: State the magnitude and a clear direction, such as east of north or below the horizontal.

Lesson summary

Choose and state the physical system, reference frame, and positive directions.
Resolve vectors into signed components and solve each perpendicular direction separately.
For ideal projectile motion, horizontal acceleration is zero and vertical acceleration is downward.
For relative motion, add velocity vectors across the named frames.
Check units, signs, direction, significant figures, and physical reasonableness.

Check your understanding

Question 1

A projectile is launched horizontally. Which statement describes its ideal motion after launch?

Its horizontal velocity is constant, and its vertical velocity changes downward.
Its vertical velocity is constant, and its horizontal velocity increases.
Both velocity components remain constant.
Its horizontal velocity decreases because gravity acts horizontally.

Show answer and explanation

Its horizontal velocity is constant, and its vertical velocity changes downward.

With air resistance ignored, gravity acts vertically. Thus horizontal acceleration is zero and vertical velocity changes downward.

Question 2

A velocity vector has components

v_x=6.0\,\mathrm{m/s}

east and

v_y=8.0\,\mathrm{m/s}

north. What is its magnitude?

$2.0\,\mathrm{m/s}$
$10.0\,\mathrm{m/s}$
$14.0\,\mathrm{m/s}$
$48\,\mathrm{m/s}$

Show answer and explanation

10.0\,\mathrm{m/s}

The components are perpendicular, so the magnitude is

\sqrt{(6.0)^2+(8.0)^2}=10.0\,\mathrm{m/s}

.

Question 3

A person walks north relative to a moving walkway. The walkway moves east relative to the ground. Which operation gives the person's velocity relative to the ground?

Subtract the two speeds as scalars.
Add the two velocity vectors, component by component.
Use only the person's northward velocity.
Multiply the two speeds.

Show answer and explanation

Add the two velocity vectors, component by component.

Relative velocities in successive frames combine as vectors. Here, the east and north components are both part of the ground-frame velocity.

Key terms

Component: The signed part of a vector along one coordinate direction.
Projectile: An object moving through the air after launch, modelled here with air resistance ignored.
Reference frame: The viewpoint or coordinate system used to describe motion.
Relative velocity: An object's velocity described from a specified reference frame.
Vector: A quantity with both magnitude and direction, such as displacement or velocity.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B2.2. 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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B2.2 · Solve projectile and relative-motion problems with two-dimensional vectors

Ontario Grade 12 Physics

SPH4U B2.2 — separating motion into perpendicular directions

What you will learn

1. Set the frame, directions, and vector components

2. Model projectile motion in two directions

3. Add velocities for relative motion

4. Check the result

Horizontal launch from a ledge

Angled launch and return to the same height

A boat crossing a river

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review