Weight and perpendicular motion components

Key idea: Projectile motion becomes manageable when you treat it as two motions sharing the same time: constant horizontal velocity and uniformly accelerated vertical motion.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Use weight as the gravitational force W = mg.
Resolve launch velocity into perpendicular components.
Explain and calculate ideal projectile motion with negligible air resistance.

Learn the idea

Big question: How can one curved flight be treated as two simple motions without splitting the object in two?

Separate directions, share time

After release, an ideal projectile has only weight acting. Gravity changes vertical velocity while horizontal velocity remains constant. The two components belong to the same object and therefore use the same time.

Resolve the initial velocity before choosing an equation. If upward is positive, vertical acceleration is −g throughout—including at the highest point where vertical velocity is briefly zero.

Check your understanding: At the top of an ideal projectile's path, which quantities are zero?

Only the vertical velocity is zero. Horizontal velocity remains constant and acceleration is still g downward.

Explain the curved path

Horizontal displacement grows in direct proportion to time, while vertical displacement contains a t² term. Combining those relations gives a parabola.

Weight W = mg points downward. It is a force, measured in newtons; mass is the amount of matter, measured in kilograms. Do not use the words interchangeably.

Check your understanding: Why does an ideal projectile not need a horizontal force to keep moving?

With no horizontal resultant force, Newton's first law says its horizontal velocity remains constant.

Scroll diagram horizontally to read all labels.

In the ideal model, horizontal velocity stays constant while gravity changes only the vertical velocity. Both component models use the same time.

Key ideas

Choose axes and a sign convention before calculating.
Do not insert the total launch speed into a vertical equation.
At the highest point vertical velocity is zero, but acceleration is still g downward.

Relationships to know

W = mg
uₓ = u cos θ
uᵧ = u sin θ
x = uₓt
vᵧ = uᵧ − gt

Follow the reasoning

Worked example

Find position and velocity during flight

Question: A ball is launched at 20 m s⁻¹, 35° above horizontal. Find its displacement and velocity after 1.5 s. Use g = 9.81 m s⁻² and neglect air resistance.

Step 1: Resolve the launch velocity
Why: Each direction needs its own initial value.
Working: ux = 20 cos35° = 16.38 m s⁻¹; uy = 20 sin35° = 11.47 m s⁻¹.
Step 2: Find displacements
Why: Both component motions have elapsed for the same 1.5 s.
Working: x = 16.38(1.5) = 24.6 m; y = 11.47(1.5) − ½(9.81)(1.5²) = 6.17 m.
Step 3: Find velocity components
Why: Horizontal velocity is constant while vertical velocity changes.
Working: vx = 16.38 m s⁻¹; vy = 11.47 − 9.81(1.5) = −3.25 m s⁻¹.

Answer: Displacement is 24.6 m horizontally and 6.17 m above launch. Velocity is (16.4 i − 3.25 j) m s⁻¹, speed 16.7 m s⁻¹ at 11.2° below horizontal.

Check: Negative vertical velocity shows the ball has passed its highest point, while positive height shows it is still above launch level.

Now try it with support

Practise with support

A ball is projected at 25 m s⁻¹ at 37° above horizontal. Find its initial components.

Hints

The angle is measured from horizontal.
Keep the vertical component positive at launch.

View the guided answer

uₓ = 25 cos 37° = 20.0 m s⁻¹ and uᵧ = 25 sin 37° = 15.0 m s⁻¹ (3 s.f.).

Your turn

Practise independently

A ball leaves a cliff horizontally at 12 m s⁻¹ and lands 1.5 s later. Find its horizontal displacement and vertical velocity just before impact.

Check your answer

Horizontal displacement = 12(1.5) = 18 m. Taking upward positive, vᵧ = 0 − 9.81(1.5) = −14.7 m s⁻¹, so the final vertical velocity is 14.7 m s⁻¹ downward.

Common mistakes and exam guidance

Watch out for

Assuming gravity reduces the horizontal velocity in the ideal model.
Saying acceleration is zero at the top because vertical velocity is momentarily zero.

In an exam

Set up separate x and y columns but use the same time in both.
State the neglected-air-resistance assumption when using constant horizontal velocity.

Put the ideas together

Exam-style practice [7 marks]

A stone is projected horizontally at 14 m s⁻¹ from a cliff 45 m high. Find the time to reach the ground, horizontal range, impact speed and angle below horizontal. Neglect air resistance and use g = 9.81 m s⁻².

Plan before you answer

Use vertical motion to find the shared time.
Use that time horizontally.
Combine final velocity components.

View the marking points and model answer

Marking points

Uses 45 = ½gt².
Obtains t = 3.03 s.
Obtains range about 42.4 m.
Uses vy = gt to get 29.7 m s⁻¹ downward.
Keeps vx = 14 m s⁻¹.
Obtains speed about 32.8 m s⁻¹.
Obtains angle about 64.8° below horizontal.

Model answer

Vertically, 45 = ½(9.81)t², so t = 3.03 s. Range = 14(3.03) = 42.4 m. At impact vx = 14 m s⁻¹ and vy = 9.81(3.03) = 29.7 m s⁻¹ downward. Hence speed = √(14² + 29.7²) = 32.8 m s⁻¹ and θ = tan⁻¹(29.7/14) = 64.8° below horizontal.

Finish from memory

Three-question recap

What force acts on an ideal projectile after release?
Check
Its weight, vertically downward.
Which quantity links the horizontal and vertical calculations?
Check
Time.
Why is acceleration non-zero at the top?
Check
Gravity still produces downward acceleration even when vertical velocity is momentarily zero.

Try this next

Use vertical motion to find time first, then substitute that time into the horizontal motion.

Continue to Potential energy in a uniform gravitational field Return to Projectile Motion

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027