An angled launch and its apex
You will be able to: Resolve launch velocity and calculate apex and landing quantities under explicit conditions.
Is a projectile at rest at the top?
A ball launches with velocity (6, 8) m/s. At its highest point it still moves horizontally at 6 m/s even though vertical velocity is zero.
A useful starting point: A horizontal launch →
Words and symbols before equations
- Launch angle θ
- Angle of initial velocity above +x, here between 0° and 90°.
- v₀ₓ and v₀ᵧ
- Initial horizontal and vertical velocity components in m/s.
- Same-height landing
- The final vertical coordinate equals the launch coordinate.
- Apex speed
- The horizontal speed at the top of an ideal two-dimensional projectile.
What this picture assumes
Launch and landing at y = 0. Constant g = 10 m/s²; no drag. Screen axis scales differ. Complementary-angle range comparison assumes the same speed and height.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- t = 0.7986 s / 1.597 s; position (4.806, 3.189) m. Velocity (6.018, 0) m/s; speed 6.018 m/s. Range 9.613 m; maximum height 3.189 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
Resolve the initial vector: v₀ₓ = v₀ cos θ and v₀ᵧ = v₀ sin θ. Horizontal motion remains x = x₀ + v₀ₓt. Vertical motion is y = y₀ + v₀ᵧt − ½gt², with vᵧ = v₀ᵧ − gt.
At the apex vᵧ = 0, giving t_top = v₀ᵧ/g. The vertical rise is v₀ᵧ²/(2g). Unless the launch is vertical, the full velocity is not zero there. Acceleration remains downward.
For return to the launch height, the nonzero root of y − y₀ = 0 is T = 2v₀ᵧ/g. Then range is v₀ₓT. The familiar R = v₀² sin(2θ)/g assumes equal launch and landing heights and no drag. With unequal heights, solve the full vertical equation instead.
A worked example, step by step
Launch with (v₀ₓ, v₀ᵧ) = (6, 8) m/s from y = 0. Find apex time, height and same-height range using g = 10 m/s².
- The top occurs at t = 8/10 = 0.8 s.
- Rise = 8²/(2 × 10) = 3.2 m.
- Same-height flight time is 2(0.8) = 1.6 s.
- Range = 6(1.6) = 9.6 m. At the top velocity is (6, 0) m/s and acceleration is (0, −10) m/s².
Zero vertical velocity at the apex is not zero total velocity.
At the apex, can speed be nonzero while acceleration is perpendicular to velocity?
Compare with an explanation
Yes. Velocity is horizontal and gravity is vertical.
Predict. Change one thing. Explain.
Keep launch speed fixed and compare complementary launch angles. Predict range and maximum height; explain why equal range does not mean equal flight time.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t = 0.7986 s / 1.597 s; position (4.806, 3.189) m. Velocity (6.018, 0) m/s; speed 6.018 m/s. Range 9.613 m; maximum height 3.189 m. Acceleration (0, −10) m/s².
Launch and landing at y = 0. Constant g = 10 m/s²; no drag. Screen axis scales differ. Complementary-angle range comparison assumes the same speed and 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 projectile starts with components (12, 16) m/s and lands at its launch height. Find top time, rise, flight time and range, using g = 10 m/s².
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Compare with the answer and four-point rubric
- 1 point: Top time = 16/10 = 1.6 s.
- 1 point: Rise = 16²/20 = 12.8 m.
- 1 point: Flight time = 3.2 s.
- 1 point: Range = 12(3.2) = 38.4 m.
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 1Which velocity component vanishes at the apex?
Vertical velocity only, unless the launch was purely vertical.
RECALL 2When is flight time twice time to apex?
When launch and landing heights match in the no-drag constant-g model.
RECALL 3How do unequal landing heights change the method?
Solve y_f = y₀ + v₀ᵧt − ½gt² for the physical positive time.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
An angled launch and its apex
- v₀ₓ = v₀ cos θ; v₀ᵧ = v₀ sin θ.
- t_top = v₀ᵧ/g; rise = v₀ᵧ²/(2g).
- T = 2v₀ᵧ/g and R = v₀ₓT only for same-height landing.
Remember: Zero vertical velocity at the apex is not zero total velocity.
Conditions: Launch and landing at y = 0. Constant g = 10 m/s²; no drag. Screen axis scales differ. Complementary-angle range comparison assumes the same speed and height.
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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