Motion in a Plane Formula Card: Your Complete Guide to Kinematics
JEE/NEET Physics · Motion in a Plane series · Part 6 of 6 · All parts →
- One card compresses six parts — vectors, both projectile types, 2-D relativity, circles
- Aₓ = A cosθ, A_y = A sinθ · R_resultant = √(Σx² + Σy²)
- Horizontal launch: t = √(2h/g), R = u√(2h/g) · Angular: T, H, R toolkit
- River: straight across for minimum time, sin⁻¹(v_s/v_b) upstream for zero drift
- a_c = v²/r — turning is acceleration
The final card of the Motion in a Plane series — six parts distilled onto one revision sheet, plus the sports, defence and space engineering that runs on these formulas every day.
- The master formula card
- The one-rule-per-part recap
- 2-D motion in the wide world
- Worked examples: the card in action
- Final practice set
- Frequently asked questions
- Recap & key takeaways
The Master Formula Card
This single table is the entire series compressed into eleven rows. If you can reconstruct each row — not memorise it, but rebuild it from the physics behind it — you are exam-ready on Motion in a Plane. Bookmark this page and return to it every revision cycle: the card is designed so that a full recap takes under three minutes.
| What | Formula / rule | Remember |
|---|---|---|
| Resolution | Aₓ = A cosθ · A_y = A sinθ | cos along, sin across |
| Resultant | √(Aₓ² + A_y²) at tan⁻¹(A_y/Aₓ) | Pythagoras back up |
| Horizontal projectile | t = √(2h/g) · R = u√(2h/g) | height alone sets time |
| Angular projectile | T = 2u sinθ/g · H = u²sin²θ/2g | vertical movie with a start |
| Range | R = u² sin2θ/g | max at 45°, twins θ & 90−θ |
| Velocity at top | u cosθ horizontal | never zero |
| Relative 2-D | v_A − v_B component-wise | subtract the seer |
| River crossing | t = d/v_b (aim ⊥ bank) | zero drift: sin⁻¹(v_s/v_b) upstream |
| Circular motion | a_c = v²/r = ω²r | ⊥ velocity, centre-pointing |
| Circle speed | v = 2πr/T | period link |
| Centripetal force | F = mv²/r | a role, not a new force |
- Range formula conditions: R = u² sin2θ/g only holds when launch and landing are at the same height. On a table or a hill, fall back to the two-movie method.
- At the top of a projectile’s path, velocity is u cosθ, never zero — only the vertical component vanishes. Acceleration stays g throughout, including at the peak.
- Two angles give the same range: θ and 90° − θ. The steeper angle buys more time and height; the flatter one more horizontal speed.
- Centripetal force is not a fresh force to draw on a free-body diagram. Tension, gravity, friction or the normal force plays that role.
- Doubling speed quadruples a_c (v² dependence); halving the radius doubles it. NEET loves this proportionality question.
The One-Rule-Per-Part Recap
Six parts, five rules — because Parts 2 and 3 share one idea. Chant them in order and the whole series unfolds:
- Part 1 — Resolve everything onto smart axes. Choose axes that simplify the problem; a well-chosen axis can turn a two-dimensional problem into two one-dimensional ones.
- Part 2 & 3 — Horizontal and vertical motion are two independent films. Gravity never calls the horizontal set; the launch speed never touches the vertical timeline. The 45° rule is just this independence finding its balance: climb and push trade off perfectly there.
- Part 4 — Subtract the observer, component-wise. Relative velocity in a plane is ordinary subtraction done separately for x and y. The river-crossing results fall straight out of it.
- Part 5 — Turning velocity is acceleration. A change in direction is a change in the velocity vector, magnitude a_c = v²/r, pointing at the centre.
2-D Motion in the Wide World
- Every artillery and missile system solves the angular-projectile equations in real time, correcting for air drag and the Earth’s rotation beneath the parabola.
- Olympic biomechanics labs tune javelin release to ~35° (drag-adjusted from 45°) and shot-put to ~38–41°: world records set by sin 2θ management.
- ISRO’s launch windows — a rocket ascending into orbit is a projectile problem chained to circular motion: this series plus the Gravitation series built every launch.
- Banked highway curves — engineers tilt roads so the horizontal component of the normal force supplies mv²/r, reducing the friction demanded: geometry acting as a safety device.
- Cricket ball-tracking (Hawk-Eye) — full 3-D vector kinematics predicting the parabolic future of the ball, adjudicating LBW within the error bars of this chapter.
- Monsoon-soaked highways — friction drops, so safe cornering speed falls with it. The v² in mv²/r is why the same curve that felt fine at 40 km/h feels dangerous at 60.
One dimension was a rail track — forward or back, no choices. Two dimensions is a field: every direction, every trade-off. The moment direction became free, physics needed vectors to speak, components to compute, and geometry to decide outcomes. Everything after this — forces, fields, orbits — happens in this wider arena.
The toolkit in one breath: resolve (Part 1), let each axis run its own movie (Parts 2–3), subtract the observer when frames disagree (Part 4), and pay v²/r whenever direction turns (Part 5). A football pass uses all five in one second — your brain approximating what these formulas compute exactly.
Picture a park seen from above: a jogger’s zigzag (vectors), a thrown frisbee (projectile), a river ferry (relative motion), a cyclist circling the fountain (UCM). One aerial photo contains the whole series — six chapters, one playground.
Worked Examples: The Card in Action
Example 1 — Twin angles. A cricketer throws at 20 m/s. At which two angles does the ball land the same distance away? Range depends on sin 2θ, and sin 2θ is unchanged when θ → 90° − θ. So 30° and 60° are range twins: R = u² sin 60°/g = 400 × (0.866)/10 ≈ 34.6 m for both. The 30° throw arrives faster; the 60° throw hangs longer — a genuine tactical choice, not just a calculation.
Example 2 — Off the cliff. A stone leaves a 20 m cliff horizontally at 15 m/s. Time of flight depends only on height: t = √(2h/g) = √(2 × 20/10) = 2 s. Range = u·t = 15 × 2 = 30 m. Notice the launch speed never entered the time calculation — the independence of axes doing all the work.
Example 3 — The river, twice. Boat speed 5 m/s, stream 3 m/s, river 60 m wide. Minimum time: aim straight across, t = 60/5 = 12 s, and drift downstream 3 × 12 = 36 m. Zero drift: aim upstream at sin⁻¹(3/5) ≈ 37°, effective crossing speed = √(5² − 3²) = 4 m/s, so t = 60/4 = 15 s. Crossing straight is faster; crossing straight across costs time — exactly the exam distinction.
Example 4 — The banked turn. A car takes a 50 m radius curve at 20 m/s. Required centripetal force = mv²/r = m × 400/50 = 8m m/s²-worth — eight newtons per kilogram of car. On a flat road, friction must supply all of it; on a banked curve, geometry shares the load, which is why mountain roads and racetracks are tilted.
Practice Set (Answers Hidden — Try First)
Attempt each one with only the master card in front of you. If a question takes more than ninety seconds, you’re not using the card — you’re re-deriving it.
(NEET-level) 3 N east + 4 N north: resultant =
(JEE Main-level) Ball off a 45 m table at 12 m/s: range (g = 10) =
(NEET-level) u = 20 m/s, θ = 45°: range =
(JEE Main-level) v = 10 m/s on r = 2.5 m: a_c =
(NEET-level) Boat 6 m/s ⊥ stream 8 m/s: resultant =
(JEE Main-level bonus) Same projectile: u = 20 m/s, θ = 45° — maximum height?
Frequently Asked Questions
Q. What’s the single most important formula in the whole series?
There isn’t one formula — there’s one method: resolve into components and treat each axis independently. Every projectile, river and relative-motion problem in JEE/NEET collapses to that move. The formulas in the card are just its most common outcomes.
Q. Why is 45° the maximum range angle?
Range = u² sin 2θ/g, and sin 2θ peaks at 1 when 2θ = 90°, i.e., θ = 45°. It’s also the balance point: below 45° you have more speed but less flight time; above it, more time but less horizontal speed.
Q. Does air resistance ever matter in these problems?
In JEE/NEET, projectiles are assumed drag-free unless stated otherwise. In the real world, drag is why javelin throwers use ~35° and not 45° — the formulas here are the ideal limit real systems correct from.
Q. Is centripetal force a new force?
No. It is the name for whatever net force points toward the centre — tension, gravity, friction, or a normal force component. Never draw it as a separate arrow on a free-body diagram.
Q. What should I study after this series?
Newton’s Laws of Motion, next. This series gave you the geometry of motion (kinematics); the laws series explains why motion changes — and the centripetal force row of your card is the bridge between the two.
- 🧠 Full-card chant: ‘resolve, release, trade, subtract, turn’.
- 🧠 Twin angles: θ and 90° − θ share a range; 45° is the champion.
- 🏠 Daily: banked curves supply mv²/r by geometry.
- 🏠 Daily: Hawk-Eye adjudicates with this chapter.
- 🎯 Exam habit: 90 seconds per card problem — if longer, re-derive instead of recall.
- 🔗 Next stop: Newton’s Laws — the why behind every row of this card.
- resolution + Pythagoras = the vector cycle
- projectiles: independence of axes does everything
- 45° max range; twins share a range
- relative motion: subtract component-wise
- circles: a_c = v²/r, turning = acceleration
- at 45°, H = R/4 — instant sanity check
- 🔁 the 11-row master card
- 🔁 one rule per part
- 🔁 2-D motion = the arena of all later physics
Key Takeaways
- Motion in a plane is one method, not many topics: resolve, run each axis independently, recombine.
- Horizontal launches are timed purely by height; angular launches trade climb against push, peaking in range at 45°.
- Relative velocity is component-wise subtraction; the river problem’s two goals (fastest vs. straightest) have different aiming strategies.
- Uniform circular motion proves that changing direction alone is acceleration — v²/r toward the centre, supplied by a real force playing the centripetal role.
- Every formula on the master card should be rebuildable, not memorised — that’s the difference between a JEE topper’s card and a crib sheet.
Quick revision
- One card compresses six parts — vectors, both projectile types, 2-D relativity, circles
- Aₓ = A cosθ, A_y = A sinθ · R_resultant = √(Σx² + Σy²)
- Horizontal launch: t = √(2h/g), R = u√(2h/g) · Angular: T, H, R toolkit
- River: straight across for minimum time, sin⁻¹(v_s/v_b) upstream for zero drift
- a_c = v²/r — turning is acceleration
- The one-rule-per-part recap
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