B2.9 Investigate and analyse projectile motion

Ontario Grade 11 Physics

Kinematics

Breaking a curved path into horizontal and vertical motion

A ball thrown through the air follows a curved path. This is projectile motion: the motion of an object that travels through the air while gravity acts on it. A projectile can move horizontally and vertically at the same time. Treating these two directions separately makes its motion easier to describe. In this lesson, the system is the moving object. We use a reference frame fixed to the ground, with right as the positive horizontal direction and up as the positive vertical direction. We assume air resistance is small enough to ignore.

What you will learn

Describe a projectile and state the conditions for the projectile-motion model.
Separate a projectile’s motion into horizontal and vertical components using a clear sign convention.
Use equations for constant horizontal velocity and constant vertical acceleration.
Plan an investigation and distinguish measured evidence from predictions.
Check calculated answers for units, direction, significant figures, and reasonableness.

1. Build the model: two directions, one motion

A scalar has magnitude only. Time, mass, and speed are scalars. A vector has both magnitude and direction. Displacement, velocity, and acceleration are vectors. For example, a velocity can be described as a speed to the right or as a horizontal and vertical component.

A component is the part of a vector in one chosen direction. For a projectile launched at speed

v

at an angle

\theta

above the horizontal, the initial horizontal velocity is

v_x=v\cos\theta

and the initial vertical velocity is

v_y=v\sin\theta

. The angle is measured from the positive horizontal direction. These relationships use right-triangle trigonometry.

In the model used here, gravity gives the projectile a constant downward acceleration. With up positive, the vertical acceleration is

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

. There is no horizontal acceleration in this model, so horizontal velocity stays constant. The horizontal and vertical motions share the same time, but changes in one direction do not change the motion in the other.

A motion sketch can show the directions and the changing vertical velocity. The horizontal component points right throughout this example. The vertical component points up before the top of the path, is zero at the top, and points down afterward.

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

Projectile model: gravity acts downward and air resistance is ignored.
Horizontal acceleration is zero; vertical acceleration is constant and downward.
Use the same elapsed time for both components of the motion.

2. Equations, signs, and a motion diagram

Displacement is the change in position. Velocity describes displacement per time, and acceleration describes change in velocity per time. For constant horizontal velocity, horizontal displacement equals horizontal velocity multiplied by time. For vertical motion with constant acceleration, use the constant-acceleration relationship that includes initial vertical velocity and vertical acceleration.

Choose and state the origin before calculating. In the sketch below, the launch point is the origin, right and up are positive, and gravity points down. The dots mark equal time intervals. Their horizontal spacing is equal because horizontal velocity is constant. Their vertical spacing changes because gravity changes vertical velocity.

A negative vertical displacement means the object is below the chosen origin. A negative vertical velocity means it is moving downward. Neither negative value means the object’s speed is negative; speed is a scalar.

If a projectile lands at the same height from which it was launched, its vertical displacement at landing is zero. If it lands at a different height, use that actual vertical displacement instead. Do not assume a same-height landing unless the situation states it.

\Delta x=v_x t,\qquad \Delta y=v_{y,i}t+\frac{1}{2}a_y t^2

Write known values with SI units before substituting.
Use signed vertical displacement and acceleration.
Check whether the landing height matches the launch height.

3. Investigate and analyse projectile motion

An investigation can test whether observations agree with the projectile-motion model. First define the system, ground-based reference frame, origin, and positive directions. Then identify what will be varied and what will be measured. For example, a class could use a ball launched horizontally from a fixed height and measure its horizontal landing distance and fall time.

In a physical investigation, measured evidence comes from observations or instruments. Students could record the launch height, use video frames to estimate flight time, and measure the landing position. They should repeat trials and record measurement uncertainty, which is the range of possible error in a measurement. Those results are evidence from the investigation; the equations give a model prediction to compare with them.

A simulation can also show predicted positions at chosen time intervals. Simulation output is not a physical measurement. Label it as simulated evidence, and compare it with the model rather than claiming that a real launch occurred. In either approach, keep a record of the procedure and the values used.

To analyse results, compare horizontal positions over equal time intervals and examine whether they change by similar amounts. Then examine vertical positions: their changes should reflect downward acceleration. A graph of horizontal position against time should be a straight trend for this model. A graph of vertical position against time curves because vertical velocity changes. Differences between observations and predictions can come from measurement limits, launch variation, or air resistance omitted from the model.

Separate physical measurements, simulation output, and model predictions.
State the procedure and record how values were obtained.
Compare observed patterns with constant horizontal velocity and downward vertical acceleration.

4. A reliable solution and final check

For each problem, identify the object, reference frame, positive directions, known values, and unknown. Split any angled launch velocity into components before using the motion equations. Solve for time using the vertical information when needed, then use that same time for horizontal motion.

Keep units in substitutions. Time is measured in seconds, position in metres, velocity in metres per second, and acceleration in metres per second squared. Round the final result to a sensible number of significant figures based on the given values. A final check should include units, sign and direction, and whether the size is physically reasonable.

The examples use the simplified model. Their results are predictions under its assumptions, not measurements from an experiment.

v_{y,f}=v_{y,i}+a_y t

The vertical motion often determines flight time.
Use that time in the horizontal relationship.
Report the direction for vector quantities and check the result against the situation.

Worked example

Horizontal launch from a ledge

A ball leaves a ledge horizontally at

3.6\ \mathrm{m/s}

. The ledge is

1.25\ \mathrm{m}

above level ground. Using the projectile model, find the flight time and horizontal distance to the landing point.

Set the system and directions
The system is the ball, and the reference frame is fixed to the ground. Take the launch point as the origin, right as positive $x$ , and up as positive $y$ . The ground is at $\Delta y=-1.25\ \mathrm{m}$ . The initial vertical velocity is zero because the ball leaves horizontally.
$v_x=3.6\ \mathrm{m/s},\quad v_{y,i}=0\ \mathrm{m/s},\quad a_y=-9.8\ \mathrm{m/s^2}$
Find the time from vertical motion
Use the vertical displacement relationship. The negative displacement gives a positive time when the downward acceleration is also negative.
$-1.25\ \mathrm{m}=(0\ \mathrm{m/s})t+\frac{1}{2}(-9.8\ \mathrm{m/s^2})t^2\quad\Rightarrow\quad t=0.505\ \mathrm{s}$
Find the horizontal distance
Horizontal velocity is constant in this model. Multiply it by the flight time. To two significant figures, the ball lands about $1.8\ \mathrm{m}$ to the right of the point below the ledge.
$\Delta x=(3.6\ \mathrm{m/s})(0.505\ \mathrm{s})=1.82\ \mathrm{m}\approx1.8\ \mathrm{m}$

Answer: The predicted flight time is

0.51\ \mathrm{s}

, and the horizontal displacement is

1.8\ \mathrm{m}

to the right.

Check: The units are seconds for time and metres for displacement. Both directions are consistent with the setup. A fall of

1.25\ \mathrm{m}

taking about half a second is reasonable; the horizontal distance is less than

2\ \mathrm{m}

at this speed.

Worked example

Launch and landing at the same height

A ball is launched at

18\ \mathrm{m/s}

at

35^\circ

above level ground and lands at its launch height. Find its flight time and horizontal range.

Define the system and directions
The system is the ball, with a ground-based reference frame. Take the launch point as the origin, right and up as positive. The vertical displacement at landing is zero. Resolve the launch velocity into horizontal and vertical components.
$v_x=(18\ \mathrm{m/s})\cos35^\circ=14.7\ \mathrm{m/s},\quad v_{y,i}=(18\ \mathrm{m/s})\sin35^\circ=10.3\ \mathrm{m/s}$
Use vertical motion to find flight time
At landing, set vertical displacement to zero. The nonzero solution gives the time when the ball returns to launch height. The zero-time solution represents the launch instant, not the landing.
$0=(10.3\ \mathrm{m/s})t+\frac{1}{2}(-9.8\ \mathrm{m/s^2})t^2\quad\Rightarrow\quad t=2.10\ \mathrm{s}$
Use the same time horizontally
The horizontal velocity remains $14.7\ \mathrm{m/s}$ . Multiplying by the flight time gives the range, meaning the horizontal distance travelled before returning to the launch height.
$\Delta x=(14.7\ \mathrm{m/s})(2.10\ \mathrm{s})=30.9\ \mathrm{m}\approx31\ \mathrm{m}$

Answer: The predicted flight time is

2.1\ \mathrm{s}

, and the horizontal range is

31\ \mathrm{m}

to the right.

Check: The units reduce to seconds and metres. The positive range points right. A launch lasting about

2.1\ \mathrm{s}

with a horizontal component near

15\ \mathrm{m/s}

gives a distance near

30\ \mathrm{m}

, so the result is reasonable.

Worked example

Position during an angled launch

A ball is launched from ground level at

20\ \mathrm{m/s}

and

30^\circ

above the horizontal. In the model, find its horizontal and vertical displacement after

1.0\ \mathrm{s}

.

Choose the system and sign convention
The system is the ball and the reference frame is fixed to the ground. Take the launch point as the origin, right as positive $x$ , and up as positive $y$ . Resolve the initial velocity before calculating displacement.
$v_x=(20\ \mathrm{m/s})\cos30^\circ=17.3\ \mathrm{m/s},\quad v_{y,i}=(20\ \mathrm{m/s})\sin30^\circ=10\ \mathrm{m/s}$
Calculate horizontal displacement
There is no horizontal acceleration in the model, so the horizontal displacement after $1.0\ \mathrm{s}$ is the horizontal velocity multiplied by time.
$\Delta x=(17.3\ \mathrm{m/s})(1.0\ \mathrm{s})=17.3\ \mathrm{m}\approx17\ \mathrm{m}$
Calculate vertical displacement
Use the initial upward velocity and the negative downward acceleration. The positive result means the ball is still above its launch point at this time.
$\Delta y=(10\ \mathrm{m/s})(1.0\ \mathrm{s})+\frac{1}{2}(-9.8\ \mathrm{m/s^2})(1.0\ \mathrm{s})^2=5.1\ \mathrm{m}$

Answer: After

1.0\ \mathrm{s}

, the ball is about

17\ \mathrm{m}

to the right and

5.1\ \mathrm{m}

above its launch point.

Check: Each displacement is in metres. The horizontal direction is right and the vertical displacement is positive, meaning up. The ball began with upward motion, so being above the launch point after one second 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 into components with trigonometry before calculating motion in each direction.

Using upward as positive but assigning gravity a positive value.

Correction: With up positive, gravity points downward, so vertical acceleration is negative.

Assuming the time for horizontal motion is different from the time for vertical motion.

Correction: Both components describe the same object during the same elapsed time.

Assuming every projectile lands at its launch height.

Correction: Use the actual landing position. Set vertical displacement to zero only when launch and landing heights match.

Calling a simulation result a measured laboratory result.

Correction: Identify whether a value came from physical measurement, simulation output, or a model prediction.

Lesson summary

Projectile motion can be analysed by separating horizontal and vertical components.
In the simplified model, horizontal velocity is constant and vertical acceleration is downward.
Choose an origin and positive directions, then use signed displacements consistently.
Use vertical information to find time when useful, and use the same time for horizontal motion.
Compare investigation evidence with model predictions, and check units, direction, significant figures, and reasonableness.

Check your understanding

Question 1

A projectile is moving upward after launch. With up chosen as positive and air resistance ignored, what is its vertical acceleration?

$+9.8\ \mathrm{m/s^2}$
$-9.8\ \mathrm{m/s^2}$
$0\ \mathrm{m/s^2}$
correctIndex":1,

Show answer and explanation

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

Gravity points downward, opposite to the positive vertical direction. Its acceleration is negative even while the projectile is moving upward.

Question 2

A projectile has a constant horizontal velocity of

4.0\ \mathrm{m/s}

. How far horizontally does it travel in

2.0\ \mathrm{s}

?

$2.0\ \mathrm{m}$
$6.0\ \mathrm{m}$
$8.0\ \mathrm{m}$
correctIndex":2

Show answer and explanation

8.0\ \mathrm{m}

Horizontal displacement is horizontal velocity multiplied by time:

(4.0\ \mathrm{m/s})(2.0\ \mathrm{s})=8.0\ \mathrm{m}

.

Key terms

Projectile: An object moving through the air while gravity acts on it.
Component: The part of a vector in one chosen direction.
Reference frame: The viewpoint and coordinate system used to describe motion.
Displacement: The change in an object's position, including direction.
Range: The horizontal distance travelled before a projectile reaches its landing point.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.9. 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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B2.9 · Investigate and analyse projectile motion

Ontario Grade 11 Physics

Breaking a curved path into horizontal and vertical motion

What you will learn

1. Build the model: two directions, one motion

2. Equations, signs, and a motion diagram

3. Investigate and analyse projectile motion

4. A reliable solution and final check

Horizontal launch from a ledge

Launch and landing at the same height

Position during an angled launch

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through SPH3U

About this lesson and its review