Angled launches, apex & unequal heights
Solve for an event vertically, then use the shared time horizontally.
A launched ball rises while moving sideways.
Need an earlier step? Start with vector components →
Launch a ball from ground level with horizontal velocity +6 m/s and vertical velocity +10 m/s. Neglect air resistance and use g = 10 m/s². After 1 s it reaches its highest point—but is still moving right.
Words and symbols you will use
- Launch components
- The initial velocity split into horizontal and vertical parts.
- Apex
- The highest point; vertical velocity is momentarily zero.
- Total speed
- The magnitude of the complete velocity: √(vₓ² + vᵧ²).
Read the picture, one step at a time.
- At 1 s: vertical velocity is 10 − 10(1) = 0 m/s and height is 5 m.
- Horizontal velocity stays +6 m/s. At the apex, total speed is therefore 6 m/s, not zero.
- At 2 s the ball returns to launch height, 12 m horizontally from the start. Its vertical velocity is −10 m/s.
| Compare | What it means | What follows |
|---|---|---|
| Vertical velocity vᵧ | Zero for an instant | The ball changes from rising to falling. |
| Horizontal velocity vₓ | +6 m/s in this example | Sideways motion continues. |
| Vertical acceleration aᵧ | −10 m/s² throughout flight | Gravity continues changing velocity. |
The idea to keep: The return time is twice the rise time only for matching launch and landing heights in this ideal model. With an elevated launch, solve for the actual landing height.
Is 45° always the best projectile launch angle?
Only for maximum horizontal range at fixed launch speed when launch and landing heights match, air resistance is negligible, and gravity is uniform.
For an angled launch, resolve initial velocity before doing kinematics: v_x0 = v₀ cos θ and v_y0 = v₀ sin θ when θ is above horizontal. The horizontal velocity stays constant. The vertical velocity decreases to zero at the apex, then becomes negative.
If landing is above or below launch, solve the vertical equation for the actual landing height. The familiar symmetric flight-time and range formulas no longer apply directly. Reject roots outside the physical interval and check that you chose the intended crossing.
A reliable approach
- Resolve velocity and state the angle reference.
- Solve the vertical equation for the requested event: apex or landing.
- Substitute that time into horizontal motion; check whether equal-height shortcuts are valid.
Work through one example.
A ball launches at 20 m/s, 30° above horizontal, and lands at its launch height. Use g = 10 m/s². Find components, flight time, maximum height above launch, and range.
Follow the worked solution
- v_x0 = 20 cos 30° = 10√3 ≈ 17.32 m/s; v_y0 = 10 m/s.
- t_apex = 10/10 = 1 s; total flight = 2 s.
- Height gain = 10²/(2·10) = 5 m; range = (10√3)(2) ≈ 34.64 m.
At the apex, only v_y is zero. Total speed is not zero for a nonvertical projectile. Do not use the equal-height range formula for a cliff launch.
Explain it without notes: Explain why complementary angles need not give equal range for a cliff launch.
Try explaining it now.
At a fixed angle and equal launch/landing heights, doubling launch speed doubles range. Agree?
Compare with an explanation
Disagree. Flight time doubles and horizontal speed doubles, so range increases by a factor of four.
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: Explain why complementary angles need not give equal range for a cliff launch.
Make a prediction. Test it.
Use a 20 m/s launch at 30° with zero launch height. Switch to 60° and compare range and peak height. Then increase launch height and explain why equal-height symmetry no longer applies.
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 projectile launches from 5 m above ground with v_x0 = 6 m/s and v_y0 = 10 m/s. Use g = 10 m/s².
- Find the apex time, height, velocity components and acceleration components.
- Derive the landing time.
- Find range.
- Explain why doubling only v_x0 changes range but not time.
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: At t = 1 s, y = 10 m; velocity is (6, 0) m/s and acceleration is (0, −10) m/s².
- 1 point: 0 = 5 + 10t − 5t² gives t = 1 + √2 ≈ 2.414 s; discard the negative root.
- 1 point: R = 6(1 + √2) m ≈ 14.49 m.
- 1 point: The vertical equation and its landing time stay unchanged; x = v_x0 t doubles when only v_x0 doubles.
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 1What is the velocity direction at the apex?
Horizontal for a nonvertical ideal projectile.
RECALL 2When do complementary launch angles give equal range?
At equal launch/landing heights, fixed speed and negligible drag.
RECALL 3What replaces the symmetric flight-time formula for unequal heights?
Solve y_f = y₀ + v_y0t − ½gt² and select the physical root.
Come back tomorrow: answer these with the cards closed. Try again a week later, especially the ones you missed.
Keep the key ideas handy.
Angled launches, apex & unequal heights
Core idea: Only for maximum horizontal range at fixed launch speed when launch and landing heights match, air resistance is negligible, and gravity is uniform.
- v_x0 = v₀ cos θ; v_y0 = v₀ sin θ
- t_apex = v_y0/g for an upward launch.
- Height gain = v_y0²/(2g)
- Same-height landing only: T = 2v_y0/g; R = v₀² sin(2θ)/g
- At apex: v = (v_x0, 0), a = (0, −g). Horizontal speed remains nonzero for nonvertical flight.
Avoid this: At the apex, only v_y is zero. Total speed is not zero for a nonvertical projectile. Do not use the equal-height range formula for a cliff launch.
Remember why: The return time is twice the rise time only for matching launch and landing heights in this ideal model. With an elevated launch, solve for the actual landing height.
Equal-height shortcuts require equal heights. Otherwise solve the actual vertical landing equation.
Explain, don’t just substituteExplain why complementary angles need not give equal range for a cliff launch.
Refresh Kid · AP Physics 1 · Unit 1 · 1.5.A, 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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