Solve constant-acceleration and free-fall problems
Choose a sign convention, select the equation that matches the known quantities, and model free fall with a uniform gravitational acceleration.
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 near Earth, which way is acceleration during vertical motion?
What you will learn
- Solve straight-line constant-acceleration problems
- Apply the equations to free fall without air resistance
- Keep signs, units and stated assumptions consistent
Choose the direction before substituting
Write down the positive direction and list known quantities with signs and units. Decide which equation contains the unknown and the known values. Check that acceleration is constant before using a standard equation.
After solving, check whether the sign and size fit the motion. A negative velocity can mean motion opposite to the chosen positive axis; it does not mean an object has a negative speed.
Free fall is a constant-acceleration model
Near Earth's surface, an object in free fall is modelled as having a uniform downward acceleration g when air resistance is neglected. Use the value of g supplied in the question; if none is supplied, 9.81 m s⁻² is a suitable illustrative value.
If upward is positive, a = −g. A ball thrown upward slows while moving up, has zero instantaneous velocity at its highest point, and then accelerates down. Its acceleration remains downward throughout in this ideal model.
Pause and explain
A ball is thrown upward and air resistance is neglected. What is its acceleration at the highest point?
Worked example
A ball is thrown vertically upward at 19.6 m s⁻¹. Take g = 9.8 m s⁻² and upward as positive. Find the time to its highest point and the upward displacement.
Show the worked solution
- At the highest point v = 0; a = −9.8 m s⁻².
- Use v = u + at: 0 = 19.6 − 9.8t, so t = 2.0 s.
- Use s = ut + ½at² = 19.6(2.0) − 4.9(4.0).
Answer Time 2.0 s; upward displacement 19.6 m.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 stone is released from rest and falls for 1.5 s. Take g = 9.8 m s⁻² and neglect air resistance. Find its speed and downward displacement.
Give me a hint
Take downward as positive and use the equations for constant acceleration.
Compare my reasoning
- u = 0, a = +9.8 m s⁻² and t = 1.5 s.
- v = u + at = 9.8 × 1.5.
- s = ut + ½at² = 0.5 × 9.8 × 1.5².
Speed 14.7 m s⁻¹; displacement 11.0 m downward.
Look for these in your work
- Initial condition and sign convention stated
- Final velocity calculated
- Displacement and direction calculated
A ball is thrown downward at 3.0 m s⁻¹ from a ledge and falls 20 m. Take downward as positive and g = 10 m s⁻². Find its speed just before impact.
Give me a hint
Use v² = u² + 2as because time is not given.
Compare my reasoning
- u = +3.0 m s⁻¹, s = +20 m and a = +10 m s⁻².
- v² = 3.0² + 2(10)(20) = 409.
- Take the positive square root for downward speed.
About 20.2 m s⁻¹ downward.
Look for these in your work
- Signed data consistent
- Correct equation selected
- Root and direction interpreted
A ball moving upward at 12 m s⁻¹ is observed 1.0 s later. Take upward as positive and g = 10 m s⁻². Find its velocity and describe its direction.
Give me a hint
Use a = −g while the ball is rising.
Compare my reasoning
- u = +12 m s⁻¹, a = −10 m s⁻² and t = 1.0 s.
- v = u + at = 12 − 10.
- Interpret the positive result as upward.
2.0 m s⁻¹ upward.
Look for these in your work
- Acceleration sign is downward
- Velocity calculated
- Direction interpreted
Common mistakes
- Using +g when upward has been chosen as positive
- Assuming acceleration is zero at the highest point
- Using a kinematics equation without checking constant acceleration
What can you explain now?
In the no-air-resistance model near Earth, which way is acceleration during vertical motion?
Compare with the explanation
Downward throughout
Gravity acts downward even while the object is travelling upward.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How does the sign convention affect free-fall calculations, and why does acceleration remain non-zero at the top?
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.