Weight and perpendicular motion
Key idea: H2 Physics lessons on weight, projectile components, gravitational potential energy, air resistance and terminal velocity.
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The core idea
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Big question: Why can horizontal and vertical motion be solved independently?
Weight is mg downward in a uniform gravitational field. With negligible air resistance, gravity changes only the vertical velocity of a projectile; horizontal velocity remains constant. Resolve the initial velocity once, use the same time for both components, and recombine them only when a final speed or direction is required.
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.
Key ideas to keep
- The two components share time but have different accelerations.
- At the highest point, vertical velocity is zero but acceleration remains g downward.
- A curved path does not require a force along the path.
See 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.
Another worked model
Question
A ball is projected horizontally at 12 m s⁻¹ from a cliff 44.1 m high. Use g = 9.8 m s⁻² and neglect air resistance. Find its time, range, impact speed and direction.
Check the worked solution
Take downward positive vertically. From 44.1 = ½(9.8)t², t = 3.00 s. Range = 12(3.00) = 36.0 m. At impact vₓ = 12 m s⁻¹ and vᵧ = 9.8(3.00) = 29.4 m s⁻¹ downward. Speed = √(12² + 29.4²) = 31.8 m s⁻¹ at tan⁻¹(29.4/12) = 67.8° below horizontal.
Use a hint if needed
Practise with support
Try this
A ball is launched at 20 m s⁻¹ and 30° above horizontal, landing at the launch height. Neglect drag and use g = 9.81 m s⁻². Find flight time and range.
Hint: Resolve the initial velocity, use vᵧ = 0 at maximum height, then double the time up.
Check your answer
uₓ = 20 cos 30° = 17.3 m s⁻¹ and uᵧ = 10.0 m s⁻¹. Flight time = 2uᵧ/g = 2.04 s. Range = uₓt = 17.3(2.04) = 35.3 m.
Now work without the hint
Practise independently
Your turn
A projectile starts with components uₓ = 18 m s⁻¹ and uᵧ = 12 m s⁻¹. After 1.5 s, find its displacement components and velocity components. Use g = 9.8 m s⁻² and neglect drag.
Check your answer
With right and up positive, x = 18(1.5) = 27 m and y = 12(1.5) − ½(9.8)(1.5²) = 6.98 m. Velocity components are vₓ = 18 m s⁻¹ and vᵧ = 12 − 9.8(1.5) = −2.70 m s⁻¹.
Avoid these traps
Common mistakes
Common mistake
Mass and weight are the same quantity.
What is wrong with this reasoning?
Show better thinking
Mass is measured in kilograms and characterises inertia. Weight is the gravitational force W = mg, measured in newtons, and changes when gravitational field strength changes.
Common mistake
Gravity gradually reduces a projectile’s horizontal velocity when air resistance is neglected.
What is wrong with this reasoning?
Show better thinking
Gravity acts vertically, so horizontal acceleration is zero and horizontal velocity is constant. The vertical component changes with acceleration g; the shared time links the two components.
Write for the examiner
Exam guidance
Draw separate horizontal and vertical data columns and declare upward or downward positive before choosing equations.
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.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
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.
Come back in three days
Check what stayed with you
Recall question 1
What force acts on an ideal projectile after release?
Check the answer
Its weight, vertically downward.
Recall question 2
Which quantity links the horizontal and vertical calculations?
Check the answer
Time.
Recall question 3
Why is acceleration non-zero at the top?
Check the answer
Gravity still produces downward acceleration even when vertical velocity is momentarily zero.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Treat horizontal and vertical motion separately, then connect them through their shared time. In the ideal model, air resistance is negligible and gravitational acceleration is uniform; questions about drag must state that a resistive force is present.
- GCE A-Level H2 PhysicsTopic 5(a) / Topic 5(b) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027