Free fall and the top of a throw
You will be able to: Apply a consistent vertical sign convention and distinguish zero velocity from zero acceleration.
What changes—and what does not—at the highest point?
Toss a ball upward at 20 m/s from y = 0. In a no-drag model with g = 10 m/s², it reaches the top at 2 s and returns to its starting height at 4 s.
A useful starting point: Derive the constant-acceleration equations →
Words and symbols before equations
- Free fall
- Motion under gravity alone, including the upward part of a throw.
- g
- Positive magnitude of gravitational acceleration; use 10 m/s² here.
- Apex
- Highest point of a trajectory.
- Upward-positive axis
- Positive y and positive vᵧ point upward; gravity gives aᵧ = −g.
What this picture assumes
Vertical launch from y = 0, returning to the same height. Up is positive; g = 10 m/s², no drag. Time stops at return to the starting height.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- t = 2 s of 4 s: y = 20 m, vᵧ = 0 m/s, aᵧ = −10 m/s². Top at 2 s; rise 20 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
After release the hand is no longer pushing. In this idealized model acceleration stays −10 m/s² while the ball rises, stops momentarily and falls. Velocity is positive on the way up and negative on the way down.
Integrating aᵧ = −g gives vᵧ = v₀ᵧ − gt and y = y₀ + v₀ᵧt − ½gt². At the apex vᵧ = 0, but acceleration is not zero. The negative acceleration changes positive velocity into negative velocity.
Choose only the time interval before a collision or catch. Return to the same height has equal speed magnitude when drag is neglected. A return to a different height requires solving the height equation; equal upward and downward times cannot be assumed then.
A worked example, step by step
A ball is thrown upward at 15 m/s. Find time to the top and rise above its release point using g = 10 m/s².
- Set upward positive: v₀ᵧ = +15 m/s and aᵧ = −10 m/s².
- At the top 0 = 15 − 10t, so t = 1.5 s.
- Δy = 15(1.5) − 5(1.5)² = 11.25 m.
- The ball is instantaneously at rest vertically, with acceleration still −10 m/s².
Zero velocity at the top does not switch off gravity.
At equal heights on ascent and descent, what is equal?
Compare with an explanation
Speed magnitude is equal in this no-drag model; vertical velocities have opposite signs.
Predict. Change one thing. Explain.
Move the time control through the apex. Identify when velocity changes sign and whether the acceleration value changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t = 2 s of 4 s: y = 20 m, vᵧ = 0 m/s, aᵧ = −10 m/s². Top at 2 s; rise 20 m.
Vertical launch from y = 0, returning to the same height. Up is positive; g = 10 m/s², no drag. Time stops at return to the starting height.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant position, velocity, acceleration or integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA ball leaves y = 0 upward at 20 m/s. With g = 10 m/s², find time to apex, apex height, return time and return velocity.
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Compare with the answer and four-point rubric
- 1 point: Apex at 20/10 = 2 s.
- 1 point: Height = 20(2) − 5(4) = 20 m.
- 1 point: Solve 0 = 20t − 5t²; the nonzero return time is 4 s.
- 1 point: vᵧ(4) = 20 − 40 = −20 m/s.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Is upward motion free fall?
Yes, if gravity is the only force after release.
RECALL 2What is acceleration at the apex?
Downward g, not zero.
RECALL 3What assumptions give equal speeds at equal heights?
Constant near-Earth gravity and negligible air resistance.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Free fall and the top of a throw
- aᵧ = −g; vᵧ = v₀ᵧ − gt.
- y = y₀ + v₀ᵧt − ½gt².
- Ideal near-Earth model: g = 10 m/s²; no air resistance.
Remember: Zero velocity at the top does not switch off gravity.
Conditions: Vertical launch from y = 0, returning to the same height. Up is positive; g = 10 m/s², no drag. Time stops at return to the starting height.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.3, objectives 1.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is Mechanics Unit 1: Kinematics. Quantitative motion problems stay in one or two dimensions; the optional spatial view clarifies vector notation. Students should study calculus alongside this course. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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