A horizontal launch
You will be able to: Separate horizontal and vertical motion and use the vertical equation to determine flight time.
Does moving sideways keep a ball from falling?
A ball rolls horizontally off a 20 m ledge at 5 m/s. In the no-drag model it reaches the ground after 2 s, 10 m from the foot of the ledge.
A useful starting point: Relative motion across a river →
Words and symbols before equations
- Projectile
- An object moving under gravity alone after launch in this model.
- Independent components
- Each component follows its own motion equation while sharing the same time.
- Initial vertical velocity
- For a horizontal launch, v₀ᵧ = 0; this does not make vertical acceleration zero.
- Range
- Horizontal displacement from launch to the selected landing surface.
What this picture assumes
Horizontal launch; y = 0 at the ground; up positive. Constant g = 10 m/s², no drag. Screen axis scales differ, so use coordinates rather than measuring screen angles.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- t = 1 s / 2 s; position (5, 15) m. Velocity (5, -10) m/s; speed 11.18 m/s. Range 10 m; maximum height 20 m. Acceleration (0, −10) m/s².
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Choose y = 0 at the ground and up positive. The horizontal equation is x = v₀ₓt because aₓ = 0. The vertical equation is y = h − ½gt² because v₀ᵧ = 0 and aᵧ = −g.
Set y = 0 to find flight time: t = √(2h/g). Then use that same t in the horizontal equation. Horizontal launch speed does not appear in the vertical equation, so changing it alone does not change fall time.
Velocity at landing has components (v₀ₓ, −gt). The magnitude is the square root of the sum of their squares. The downward component grows while the horizontal component stays constant. This independence depends on uniform gravity and neglected drag; it is a model, not a promise for every real object.
A worked example, step by step
A ball leaves a 5 m high table horizontally at 3 m/s. Use g = 10 m/s² to find time, range and landing velocity.
- Vertical motion gives 0 = 5 − 5t², so t = 1 s.
- Horizontal displacement is 3(1) = 3 m.
- Landing components are vₓ = 3 m/s and vᵧ = −10 m/s.
- Speed is √109 m/s ≈ 10.44 m/s, directed right and downward.
A horizontal launch has zero initial vertical velocity, not zero vertical acceleration.
A dropped ball and a horizontally launched ball leave the same height together. Which lands first?
Compare with an explanation
They land together in the ideal model if their initial vertical velocities match and drag is negligible.
Predict. Change one thing. Explain.
Hold height fixed and double horizontal launch speed. Predict flight time and range. Move the time control to the same flight fraction in both cases.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t = 1 s / 2 s; position (5, 15) m. Velocity (5, -10) m/s; speed 11.18 m/s. Range 10 m; maximum height 20 m. Acceleration (0, −10) m/s².
Horizontal launch; y = 0 at the ground; up positive. Constant g = 10 m/s², no drag. Screen axis scales differ, so use coordinates rather than measuring screen angles.
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 launches horizontally at 8 m/s from 20 m above the ground. Calculate time, range, vertical landing velocity and landing speed with g = 10 m/s².
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Compare with the answer and four-point rubric
- 1 point: Time = √(40/10) = 2 s.
- 1 point: Range = 8(2) = 16 m.
- 1 point: vᵧ = −10(2) = −20 m/s.
- 1 point: Speed = √(8² + 20²) = √464 ≈ 21.54 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 1What determines the time for a horizontal launch?
Initial height and gravitational acceleration in this model.
RECALL 2Which velocity component stays constant?
Horizontal velocity, because aₓ = 0.
RECALL 3What stays common to both component equations?
Elapsed time from launch.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A horizontal launch
- x = x₀ + v₀ₓt; y = h − ½gt².
- Flight time to ground: √(2h/g).
- a = (0, −g); use g = 10 m/s² here.
Remember: A horizontal launch has zero initial vertical velocity, not zero vertical acceleration.
Conditions: Horizontal launch; y = 0 at the ground; up positive. Constant g = 10 m/s², no drag. Screen axis scales differ, so use coordinates rather than measuring screen angles.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.5, objectives 1.5.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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