Horizontal launches: time, range & impact
Find time vertically, then range horizontally.
A ball leaves the edge of a shelf.
Need an earlier step? Start with independent horizontal & vertical motion →
A ball leaves a 5 m shelf horizontally at 3 m/s. It has no initial vertical velocity, but gravity begins changing its vertical velocity immediately. Neglect air resistance and use g = 10 m/s².
Words and symbols you will use
- Height H
- Vertical distance from the launch point down to the landing level; here 5 m.
- Range R
- Horizontal displacement from launch to landing; here motion is rightward.
- Horizontal launch
- Initial vertical velocity is zero. Vertical acceleration is not zero.
Read the picture, one step at a time.
- Use falling to find the clock: 5 m = ½(10 m/s²)t², so t = 1 s.
- Use that same second for horizontal motion: R = (3 m/s)(1 s) = 3 m.
- At landing, vₓ = +3 m/s and vᵧ = −10 m/s. The velocity is no longer horizontal.
| Compare | What it means | What follows |
|---|---|---|
| 1. How long in the air? | Use the vertical drop. | t = √(2H/g) = 1 s. |
| 2. How far horizontally? | Use the shared flight time. | R = vₓt = 3 m. |
The idea to keep: The square-root flight-time formula here assumes zero initial vertical velocity. For an upward or downward launch, return to the general vertical-position equation.
How do you solve a horizontal projectile problem?
Find fall time from the vertical drop, then multiply by horizontal velocity to find range. Combine velocity components if impact speed is requested.
A horizontal launch has zero initial vertical velocity, not zero vertical acceleration. The projectile immediately begins acquiring downward velocity. Its horizontal motion stays uniform under the ideal model.
A useful experiment varies launch speed while holding table height fixed. Measure horizontal travel from the vertical projection of the launch point, not from the edge of a ramp or an arbitrary point. Use a safe low-height setup, repeated trials and clear impact marks to reduce uncertainty.
A reliable approach
- Set y = 0 at the ground and y₀ = H at launch.
- Solve 0 = H − ½gt² for the positive time.
- Use R = v_x0t and v_y = −gt; combine components for speed.
Work through one example.
A ball leaves a 1.25 m-high table horizontally at 4 m/s. Neglect drag and use g = 10 m/s². Find time, range, and impact speed.
Follow the worked solution
- t = √(2·1.25/10) = 0.50 s.
- R = 4(0.50) = 2.0 m.
- v_y = −5 m/s, so impact speed = √(4² + 5²) = √41 ≈ 6.40 m/s.
The fall time is not H/g; that expression has the wrong units. It is √(2H/g) for release with zero vertical velocity.
Explain it without notes: Describe a measurement that could test the predicted square-root dependence.
Try explaining it now.
You double launch height while holding horizontal speed fixed. Does range double?
Compare with an explanation
No. With zero initial vertical velocity, flight time and range both scale with √H, so each grows by √2.
Another explanation, if you need oneOptional external lesson · Flipping Physics
Two-dimensional and projectile motion
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 16:57–20:13. 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: Describe a measurement that could test the predicted square-root dependence.
Make a prediction. Test it.
Set height to 1.25 m and horizontal speed to 4 m/s. Check the numerical results, then quadruple height and predict the new time and range.
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 simulated horizontal launcher has fixed unknown speed u. Ideal measurements give (H, R) = (0.80 m, 1.20 m), (1.80 m, 1.80 m), (3.20 m, 2.40 m). Use g = 10 m/s².
- Derive a linear relation between R² and H.
- Use the data to find its slope and u.
- Describe measurements and controls for testing this with a small classroom launcher.
- Identify a realistic limitation.
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: t² = 2H/g and R = ut give R² = (2u²/g)H.
- 1 point: Slope ΔR²/ΔH = (5.76 − 1.44)/(3.20 − 0.80) = 1.8 m; u = √(g·slope/2) = 3.0 m/s.
- 1 point: Measure launch height relative to the landing surface and range from the launch point’s vertical projection. Keep launch speed and horizontal release fixed; repeat at several heights.
- 1 point: Examples: release angle/speed changes or uncertainty in locating the impact point. The ideal dataset here has no measurement noise; real data do.
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 1Does v_y0 = 0 imply a_y = 0?
No. a_y = −g immediately after launch.
RECALL 2How does doubling height affect time?
Time increases by √2 when other vertical conditions match.
RECALL 3How does doubling horizontal speed affect range?
It doubles the range without changing fall time.
Come back tomorrow: answer these with the cards closed. Try again a week later, especially the ones you missed.
Keep the key ideas handy.
Horizontal launches: time, range & impact
Core idea: Find fall time from the vertical drop, then multiply by horizontal velocity to find range. Combine velocity components if impact speed is requested.
- t_fall = √(2H/g)
- R = v_x0√(2H/g)
- Impact speed = √(v_x0² + 2gH)
- At fixed u: R² = (2u²/g)H; slope of R² vs H has units m.
Avoid this: The fall time is not H/g; that expression has the wrong units. It is √(2H/g) for release with zero vertical velocity.
Remember why: The square-root flight-time formula here assumes zero initial vertical velocity. For an upward or downward launch, return to the general vertical-position equation.
Horizontal release: v_y0 = 0; level landing surface; no drag.
Explain, don’t just substituteDescribe a measurement that could test the predicted square-root dependence.
Refresh Kid · AP Physics 1 · Unit 1 · 1.5.B · 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 1.5. 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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