Model horizontal motion with vertical acceleration
Treat perpendicular horizontal and vertical motion independently while using one shared time interval.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 2. The falling-object activity is an online description and calculation practice; it does not provide supervised laboratory competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use SI units and signed values, and recall that velocity includes direction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In the no-air-resistance model, what is the horizontal acceleration of a horizontally launched object?
What you will learn
- Describe uniform velocity in one direction and uniform acceleration perpendicular to it
- Resolve motion into horizontal and vertical components
- Calculate a trajectory while stating the no-air-resistance assumption
Perpendicular components can be analysed separately
When an object has uniform horizontal velocity and uniform vertical acceleration, treat the two perpendicular components as separate one-dimensional motions. In the ideal projectile model with air resistance neglected, horizontal acceleration is zero and vertical acceleration is downward g.
Both components share the same elapsed time. Horizontal displacement is x = uₓt. If the object is launched horizontally, initial vertical velocity is zero and vertical displacement is y = ½gt² downward. Combine the component displacements only after finding the common time.
The path curves even though each component is simple
The horizontal velocity stays constant while the vertical velocity changes uniformly. Their combined velocity changes direction, so the path is curved. At any instant, the velocity is the vector combination of horizontal and vertical components.
The model assumes a uniform gravitational field and neglects air resistance. It is suitable for explaining the specified motion pattern, but real trajectories can differ when drag or other forces matter.
Pause and explain
How are horizontal and vertical motions linked in a projectile calculation?
Worked example
A ball is projected horizontally at 12 m s⁻¹ from a 45 m ledge. Take g = 9.8 m s⁻² and neglect air resistance. Find its flight time and horizontal range.
Show the worked solution
- Vertical motion: 45 = ½(9.8)t², so t = √(90/9.8) ≈ 3.03 s.
- Horizontal velocity is constant, so range = 12 × 3.03.
- State that the result uses the no-air-resistance model.
Answer Flight time ≈ 3.03 s; horizontal range ≈ 36.4 m.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A ball leaves a horizontal table at 8.0 m s⁻¹ and falls 1.25 m. Take g = 10 m s⁻². Find the time to fall and horizontal distance.
Give me a hint
Find time from vertical motion first, then use that same time horizontally.
Compare my reasoning
- 1.25 = ½(10)t², so t² = 0.25 and t = 0.50 s.
- Horizontal displacement = 8.0 × 0.50.
- Neglect air resistance as stated by the model.
Time 0.50 s; horizontal distance 4.0 m.
Look for these in your work
- Vertical equation used for time
- Shared time used horizontally
- Distance calculated
A projectile has horizontal velocity 6.0 m s⁻¹ and vertical velocity 8.0 m s⁻¹ at one instant. Find its speed and direction above the horizontal.
Give me a hint
Combine perpendicular velocity components.
Compare my reasoning
- Speed = √(6.0² + 8.0²).
- Angle = tan⁻¹(8.0/6.0).
- Report the direction above the horizontal.
Speed 10 m s⁻¹ at about 53° above the horizontal.
Look for these in your work
- Perpendicular components combined
- Magnitude and angle calculated
A package is released horizontally at 15 m s⁻¹ from a height of 80 m. Take g = 10 m s⁻² and neglect air resistance. Find the flight time and horizontal distance.
Give me a hint
Use vertical fall to determine time, then constant horizontal velocity.
Compare my reasoning
- 80 = ½(10)t², so t² = 16 and t = 4.0 s.
- Horizontal distance = 15 × 4.0.
- Keep the assumptions and units with the result.
Flight time 4.0 s; horizontal distance 60 m.
Look for these in your work
- Vertical displacement used to find time
- Horizontal motion treated as uniform
- Both results include units
Common mistakes
- Using the horizontal speed to calculate vertical fall time
- Giving horizontal motion a gravitational acceleration
- Forgetting that both component motions use the same time
What can you explain now?
In the no-air-resistance model, what is the horizontal acceleration of a horizontally launched object?
Compare with the explanation
Zero
Gravity acts vertically, so horizontal velocity is uniform in this model.
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Make it stick
Why can horizontal and vertical motion be solved separately, and what connects the two calculations?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.