Free fall & vertical launches
Describe the apex without confusing zero velocity with zero acceleration.
Follow a ball after it leaves your hand.
Need an earlier step? Start with acceleration, signs & changing speed →
Throw a ball straight up at 10 m/s from height 0. After release, neglect air resistance and use g = 10 m/s². Choose up as positive. Gravity changes the ball’s velocity by −10 m/s each second.
Words and symbols you will use
- Free fall
- Motion during which gravity is the only force being modeled; it can include upward motion.
- vᵧ
- Vertical velocity. Positive is upward in this example.
- g
- The positive magnitude of gravitational acceleration. Here g = 10 m/s², so aᵧ = −10 m/s².
Read the picture, one step at a time.
- At release: vᵧ = +10 m/s. The ball moves upward but accelerates downward.
- At 1 s: vᵧ = 0 and the height is 5 m. Acceleration is still −10 m/s².
- Just before returning at 2 s: vᵧ = −10 m/s. The ball moves downward and is speeding up.
| Compare | What it means | What follows |
|---|---|---|
| Going up | vᵧ positive | aᵧ = −10 m/s²; speed decreases. |
| Highest point | vᵧ = 0 for an instant | aᵧ = −10 m/s²; no pause in gravity. |
| Coming down | vᵧ negative | aᵧ = −10 m/s²; speed increases. |
The idea to keep: Zero velocity means “not changing position at that instant.” Zero acceleration would mean “not changing velocity.” They are different claims. Contact with a hand or the ground is outside this flight model.
Is acceleration zero at the top of a thrown ball’s path?
No. At the highest point, vertical velocity is momentarily zero, but gravitational acceleration remains downward.
Free fall means gravity is the only force being modeled during the flight. A ball thrown upward is in free fall after release if air resistance is neglected. Its upward motion does not require upward acceleration.
Choose up as positive. Then a_y = −g throughout the airborne interval. On the way up, positive velocity decreases. On the way down, negative velocity becomes more negative. The ball has zero vertical velocity only for an instant at the top; it does not wait there.
A reliable approach
- State the origin, upward-positive axis, and a_y = −g.
- Use constant-acceleration equations only during the airborne interval.
- At a peak set v_y = 0; at a landing set y equal to the landing height.
Work through one example.
A ball is launched vertically upward at 20 m/s from y = 0. Neglect air resistance and use g = 10 m/s². Find peak time, peak height, and return time to y = 0.
Follow the worked solution
- 0 = 20 − 10t gives t_peak = 2 s.
- Δy = 20(2) − 5(2²) = 20 m.
- 0 = 20t − 5t² gives t = 0 or 4 s; return time is 4 s.
Zero velocity at one instant does not imply zero acceleration. Equal up/down flight times apply only to matching launch and landing heights under this model.
Explain it without notes: Explain the difference between being momentarily at rest and staying at rest.
Try explaining it now.
At the top of an upward toss, does the ball stop accelerating?
Compare with an explanation
No. Vertical velocity is momentarily zero; acceleration remains downward. The velocity continues changing through the apex.
Another explanation, if you need oneOptional external lesson · Flipping Physics
Free-fall refresher
This lesson is complete without a video. For another teacher’s explanation, open the original resource below. Pause after a diagram and explain the idea in your own words.
Suggested section 8:06–9:06. Video by Flipping Physics / Jonathan Thomas-Palmer. Refresh Kid is not affiliated with or endorsed by Flipping Physics. These links open another website. Some videos use g = 9.81 m/s²; our examples state g = 10 m/s². Follow the value given in each problem.
After watching: Explain the difference between being momentarily at rest and staying at rest.
Make a prediction. Test it.
Select vertical speed 20 m/s and height 0 m. Scrub to the top. The vertical velocity readout reaches zero but the acceleration remains −10 m/s².
Motion graphs · same clock, different quantities
Try two questions.
Two original Refresh Kid questions. Choose an answer, explain it to yourself, then check the reasoning. These are not released AP exam questions.
Show your reasoning.
Original mini-FRQ · 4 points · Self-checkA ball is thrown upward at 15 m/s from a platform 20 m above the ground. Use g = 10 m/s² and y = 0 at ground.
- Write y(t).
- Find time to peak.
- Find peak height above ground.
- Find when it reaches ground.
Your response stays on this page and is not submitted or automatically graded. Copy it before leaving.
Compare with the worked solution & scoring guide
- 1 point: y = 20 + 15t − 5t², in SI units.
- 1 point: t_peak = 1.5 s.
- 1 point: y_peak = 31.25 m.
- 1 point: 0 = 20 + 15t − 5t² gives t = 4 s; discard t = −1 s.
Accept an equivalent correct method. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Is an upward-moving released ball in free fall?
Yes, when only gravity is modeled.
RECALL 2Does mass change free-fall acceleration in this ideal model?
No.
RECALL 3Which g value is used here?
10 m/s² in examples and simulations; follow the value stated in each problem.
Come back tomorrow: answer these with the cards closed. Try again a week later, especially the ones you missed.
Keep the key ideas handy.
Free fall & vertical launches
Core idea: No. At the highest point, vertical velocity is momentarily zero, but gravitational acceleration remains downward.
- v_y = v_y0 − gt
- y = y₀ + v_y0t − ½gt²
- At peak: v_y = 0, a_y = −g.
Avoid this: Zero velocity at one instant does not imply zero acceleration. Equal up/down flight times apply only to matching launch and landing heights under this model.
Remember why: Zero velocity means “not changing position at that instant.” Zero acceleration would mean “not changing velocity.” They are different claims. Contact with a hand or the ground is outside this flight model.
Airborne motion only; uniform gravity and negligible air resistance.
Explain, don’t just substituteExplain the difference between being momentarily at rest and staying at rest.
Refresh Kid · AP Physics 1 · Unit 1 · 1.3.A · Check units, direction and model assumptions.
Connect to released AP practice.
2026 · Question 1 · Version J
Use Part A(i) for component velocity graphs and Part A(ii) for a kinematics derivation. The remaining parts use fluid concepts from Unit 8; they are not Unit 1-only practice.
This selection directly connects to this lesson. Historical papers may use different timing or course coverage. Official questions remain on College Board’s site; our questions below are original practice.
Browse released years and scoring information ↗Framework alignment: College Board CED, Topic 1.3. Current exam corrections. Checked September 16, 2026. These explanations and practice items are independently authored by Refresh Kid. The simulation is a mathematical model, not experimental data.
About the videos and learning approach
Optional videos are linked to the publisher’s website and YouTube channel with credit. No external video player is loaded on this lesson page. The written lessons, simulations and practice here are independently authored by Refresh Kid.
We combine worked examples, visual models, explanation and recall practice. See the Institute of Education Sciences study guide ↗ for the underlying learning recommendations.
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