In two-dimensional motion, the components along perpendicular axes evolve independently under the corresponding components of acceleration. This decoupling is the key insight that reduces 2D motion to two 1D problems.
-- NCERT Class 11 Physics, Ch. 3, p. 38Motion in a Plane
8 MCQs3 revision cards9-step worked example
Source: NCERT KinematicsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Motion in a Plane, explained for NEET
Motion in a plane combines two independent one-dimensional motions — horizontal and vertical — into a single framework. The core insight: a projectile's horizontal velocity stays constant (no horizontal force, air resistance neglected), while its vertical velocity changes under gravity alone (NCERT Class 11 Physics Chapter 3, page 38). This decomposition is the engine behind every projectile and circular-motion problem NEET asks.
Projectile traps that cost marks.
The height formula H = v₀² sin²θ / (2g) and the range formula R = v₀² sin(2θ) / g share v₀ and θ but differ in the sin-vs-sin² factor and the denominator. A common confusion: plugging sin(2θ) when the question asks for height, or sin²θ when it asks for range. At θ = 45°, both formulas give the same numerator — that's the checkpoint to verify you picked the right one.
At the highest point of a projectile's trajectory, the vertical component vanishes. The speed there equals v₀ cosθ — not zero, not v₀ sinθ. Watch the angle reference: "60° with the vertical" means 30° with the horizontal. The horizontal component switches from cosθ to sinθ when the reference axis flips.
Uniform circular motion (UCM) — the velocity-vs-speed distinction.
In UCM, speed is constant but velocity is not. Direction changes continuously, producing centripetal acceleration a_c = v²/r directed inward. A question asking "which quantity is constant?" tests whether you treat velocity as a scalar (wrong) or a vector (correct). Speed and kinetic energy are constant; velocity and acceleration direction are not.
The UCM-to-projectile bridge.
A particle completing circular motion at radius R with period T has speed v = 2πR/T. If that particle is then launched vertically, the max height is H = v²/(2g) = (2πR/T)²/(2g). The trap: using R directly as the height.
Relative velocity in 2D is vector subtraction — same-direction objects have relative velocity v_A − v_B (smaller magnitude); opposite-direction objects have v_A + v_B (larger magnitude). Sign errors here are common under time pressure.
Can you answer these Motion in a Plane MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPYQ Pattern
A projectile is launched with speed 20 m/s at 30° above the horizontal. What is the maximum height reached? (Take g = 10 m/s²)
Show answer and why every option is right or wrong
Answer: B. H = v₀² sin²θ / (2g) = (20)² × sin²(30°) / (2 × 10) = 400 × 0.25 / 20 = 5.0 m. Here g = 10 m/s² is a problem-defined exact value and 30° is an exact angle — they do not limit significant figures. (NCERT Class 11 Physics Chapter 3, page 40.)
Why A is wrong: A gives 10 m — this results from using sinθ instead of sin²θ in the height formula: 400 × 0.5 / 20 = 10. The height formula requires sin squared. (trap: height vs range formula confusion)
Why C is wrong: C gives 20 m — this results from using the range formula R = v₀² sin(2θ)/g = 400 × sin(60°)/10 ≈ 34.6, or from omitting the factor of 2 in the denominator: 400 × 0.25 / 10 = 10, then doubling incorrectly. Neither matches the height formula.
Why D is wrong: D gives 15 m — this results from using sin²(60°) instead of sin²(30°): 400 × 0.75 / 20 = 15. The launch angle is 30° above horizontal, not 60°. (trap: angle reference confusion)
MCQ 2Concept TrapPYQ Pattern
A projectile is launched at 60° with the vertical. What is its launch angle with respect to the horizontal, and what is its speed at the highest point in terms of the launch speed v₀?
Show answer and why every option is right or wrong
Answer: A. A is correct. The vertical and the horizontal are 90° apart, so an angle of 60° from the vertical is 90° − 60° = 30° from the horizontal. At the highest point only the horizontal component of velocity survives — the vertical component is momentarily zero — so the speed there is v₀ cos θ measured from the HORIZONTAL, which is v₀ cos 30° = v₀√3/2 ≈ 0.87 v₀. Both halves of this question turn on the same habit: convert to the horizontal reference first, then apply the formula, because every projectile formula is written in terms of the angle from the horizontal (NCERT Class 11 Physics Chapter 3, page 40).
Why B is wrong: B is wrong on both counts, and for one reason: it reports 60° straight from the stem without converting. 60° is the angle from the VERTICAL; from the horizontal it is 30°. Feeding the unconverted 60° into v₀ cos θ then gives v₀/2 as well. (trap: angle reference axis confusion)
Why C is wrong: C is wrong because 45° would require the angle from the vertical to be 45° too, which contradicts the stated 60°. A projectile is only symmetric about the two axes at 45°.
Why D is wrong: D has the angle right but the speed wrong. Having converted correctly to 30° from the horizontal, it then evaluates cos 30° as 1/2. cos 30° = √3/2 ≈ 0.87; it is cos 60° that equals 1/2.
MCQ 3Easy RecallPYQ Pattern
In uniform circular motion at constant speed, which of the following is constant?
Show answer and why every option is right or wrong
Answer: A. In UCM, speed is constant, so KE = ½mv² is constant. Velocity (a vector) changes direction continuously. Acceleration has constant magnitude v²/r but its direction (always toward the centre) rotates — so the acceleration vector is not constant. (NCERT Class 11 Physics Chapter 3, page 43.)
Why B is wrong: B is wrong — centripetal acceleration has constant magnitude (v²/r) but its direction rotates with the object, always pointing toward the centre. The acceleration vector is therefore not constant. (trap: treating acceleration as scalar)
Why C is wrong: C is wrong — velocity is a vector. In UCM the direction of motion changes continuously, so velocity is not constant even though its magnitude (speed) is. (trap: confusing speed with velocity)
Why D is wrong: D is wrong — centripetal force always points toward the centre, but the object moves around the circle, so the direction of this force (relative to a fixed coordinate system) rotates continuously.
MCQ 4Direct ApplicationPractice
A particle moves in a circle of radius 0.50 m with a constant speed of 2.0 m/s. The magnitude of centripetal acceleration is:
Show answer and why every option is right or wrong
Answer: D. a_c = v²/r = (2.0)²/0.50 = 4.0/0.50 = 8.0 m/s². (NCERT Class 11 Physics Chapter 3, page 43.)
Why A is wrong: A gives 1.0 m/s² — this comes from v/r² = 2.0/0.25 = 8.0 (not 1.0) or from v·r = 1.0. Neither is the centripetal acceleration formula.
Why B is wrong: B gives 4.0 m/s² — this is v² = 4.0, forgetting to divide by r. The formula is a_c = v²/r, not v².
Why C is wrong: C gives 2.0 m/s² — this is v/r = 2.0/0.50 = 4.0 (not even 2.0), or simply the speed itself. Centripetal acceleration is v²/r, not v/r.
MCQ 5Easy RecallPractice
A body moving in a circle at constant speed has:
Show answer and why every option is right or wrong
Answer: C. In uniform circular motion, the acceleration is centripetal — directed radially inward toward the centre. Speed is constant but velocity direction changes, producing this inward acceleration of magnitude v²/r. (NCERT Class 11 Physics Chapter 3, page 43.)
Why A is wrong: A is wrong — 'speed constant' does not mean 'acceleration zero.' Acceleration is the rate of change of velocity (a vector). In UCM, the velocity direction changes continuously, so acceleration is non-zero. (trap: treating velocity as a scalar / mistake: claiming zero acceleration in UCM)
Why B is wrong: B is wrong — if acceleration were along the velocity, it would change the speed (speed up or slow down the object). In UCM, speed is constant, so there is no tangential component of acceleration.
Why D is wrong: D is wrong — acceleration directed away from the centre describes centrifugal effects in a rotating frame. In an inertial frame, the acceleration is centripetal (inward).
MCQ 6Easy RecallPractice
A projectile is launched at speed v₀ at angle θ above the horizontal. Which expression gives the horizontal range on level ground?
Show answer and why every option is right or wrong
Answer: C. The horizontal range for a projectile returning to the same level is R = v₀² sin(2θ)/g. This is derived from the time of flight T = 2v₀ sinθ/g and horizontal distance R = v₀ cosθ × T. (NCERT Class 11 Physics Chapter 3, page 40.)
Why A is wrong: A gives the maximum height formula H = v₀² sin²θ/(2g), not the range. The two formulas share v₀ and θ but differ: height has sin² and 2g in the denominator, range has sin(2θ) and g alone. (trap: height vs range formula swap)
Why B is wrong: B gives v₀² cosθ/g — this is dimensionally consistent but is not a standard projectile formula. It confuses the horizontal component v₀ cosθ with the full range expression.
Why D is wrong: D gives 2v₀² sinθ/g — this equals v₀ times the time of flight (2v₀ sinθ/g), not the range. The range uses the horizontal component v₀ cosθ multiplied by the time of flight, yielding sin(2θ) via the identity 2 sinθ cosθ = sin(2θ).
MCQ 7Direct ApplicationPractice
A ball is thrown at 37° above the horizontal with speed 10 m/s. (Take g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8.) The total time of flight is:
Show answer and why every option is right or wrong
Answer: B. The time of flight for a projectile launched and landing at the same level is T_f = 2v₀ sinθ/g = 2 × 10 × 0.6 / 10 = 1.2 s. (NCERT Class 11 Physics, Chapter 3, page 40.)
Why A is wrong: A gives 0.6 s — this is v₀ sinθ/g = 10 × 0.6/10 = 0.6, the time to reach maximum height (t_m), not the total time of flight. The upward and downward halves of the flight take equal time, so T_f is twice t_m.
Why C is wrong: C gives 1.6 s — this results from using cosθ in place of sinθ: 2 × 10 × 0.8/10 = 1.6. The time-of-flight formula depends on the vertical velocity component (v₀ sinθ), not the horizontal component.
Why D is wrong: D gives 2.0 s — this comes from omitting the sinθ factor entirely: 2v₀/g = 2 × 10/10 = 2.0. The formula requires the vertical component v₀ sinθ, not v₀ alone.
MCQ 8CalculationPYQ Pattern
Two buses travel a route of length L. Bus A and Bus B move in opposite directions, each at speed v_b. A car travels the route at speed v_c in the same direction as Bus A. If buses from each end depart every T minutes, the car encounters an oncoming bus (Bus B) every T_opp = L/(v_b + v_c) minutes and a same-direction bus (Bus A) overtakes or is overtaken every T_same = L/(v_b − v_c) minutes. If T_opp = 5 min and T_same = 15 min, the ratio v_b/v_c is:
Show answer and why every option is right or wrong
Answer: B. From the two equations: (v_b + v_c)/(v_b − v_c) = T_same/T_opp = 15/5 = 3. So v_b + v_c = 3v_b − 3v_c → 4v_c = 2v_b → v_b/v_c = 2. The ratio is 2:1.
Why A is wrong: A gives 1:1 — if v_b = v_c, the same-direction bus would never pass the car (denominator v_b − v_c = 0), making T_same infinite. This contradicts T_same = 15 min. (trap: ignoring direction in relative velocity)
Why C is wrong: C gives 3:1 — this comes from setting (v_b + v_c)/(v_b − v_c) = 3 but then solving incorrectly, e.g., assuming v_b = 3v_c directly without expanding.
Why D is wrong: D gives 4:1 — this results from an arithmetic error such as 4v_c = 2v_b → v_b = 4v_c (dividing wrong side). Correctly: 2v_b = 4v_c → v_b = 2v_c.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Motion in a Plane: quick recall before you leave
How do you solve a Motion in a Plane question? A worked example
Pattern: Projectile launched from UCM — find maximum height (pattern: projectile from circular motion).
- 1
Given
A particle moves in uniform circular motion with radius R = 0.20 m and period T = 0.40 s. It is then launched vertically upward with the same speed it had in circular motion. Take g = 9.8 m/s², π² = 9.87.
- 2
Required
Maximum height H reached by the particle after vertical launch.
- 3
Concept
The particle's speed in UCM is v = 2πR/T. Once launched vertically, it becomes a projectile with initial upward speed v and deceleration g. At the peak, all kinetic energy converts to potential energy: H = v²/(2g).
- 4
Formula
v = 2πR/T
H = v²/(2g) - 5
Substitution
v = 2π(0.20)/0.40 = π m/s
H = (π)²/(2 × 9.8) = π²/19.6 - 6
Calculation
H = 9.87/19.6 = 0.5036 m
Note on exact constants: the factor 2 in the denominator and the formula structure (2πR/T) are exact mathematical relations. Only R, T, and g carry measurement precision. With R (2 sig figs), T (2 sig figs), and g (2 sig figs from 9.8), the result is limited to 2 significant figures. - 7
Final answer
H ≈ 0.50 m (2 significant figures)
- 8
Common trap
Using R = 0.20 m directly as the height (the UCM-to-projectile bridge trap). The radius has no direct geometric relationship to the projectile's maximum height — the speed must be computed first from v = 2πR/T.
- 9
Similar NEET-style question
A satellite moves in a circular orbit of radius 6400 km with period 90 minutes. If an object at the same speed were launched vertically from the surface, what maximum height would it reach? (Same two-step pattern: compute v from UCM, then apply H = v²/(2g).)
---
What to remember before solving Motion in a Plane questions
Which Motion in a Plane formulas do you need for NEET?
5 formulas — click to collapse
First kinematic equation (uniform acceleration)
Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant (uniform) acceleration | m/s^2 |
| t | Elapsed time | s |
Valid when
- Acceleration a is CONSTANT (uniform) in both magnitude and direction
- All quantities measured in the same inertial reference frame
- Motion is along a straight line; signs encode direction along chosen axis
Do NOT use when
- Acceleration changes in magnitude or direction (use a(t) integration)
- Motion is uniformly circular at constant speed (a is centripetal, not tangential)
Second kinematic equation (displacement under uniform acceleration)
Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.
| Symbol | Quantity | SI Unit |
|---|---|---|
| x | Final position | m |
| x0 | Initial position | m |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| t | Time elapsed | s |
Valid when
- Acceleration constant (magnitude and direction)
- Sign convention consistent across x, v, a (one chosen positive direction)
Third kinematic equation (velocity-squared)
Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| x - x0 | Displacement | m |
Valid when
- Constant acceleration
- Use signed values for v, v0, a, and (x - x0) consistently
Do NOT use when
- Time-dependent acceleration
- Curvilinear motion where acceleration is not parallel to displacement
Projectile maximum height
Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H | Maximum height (above launch) | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle | rad/deg |
| g | Gravitational acceleration | m/s^2 |
Valid when
- Air resistance neglected
- Constant g over trajectory
Projectile horizontal range
For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | Horizontal range | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle above horizontal | rad (or deg with sin in deg) |
| g | Gravitational acceleration | m/s^2 |
Valid when
- Launch and landing are at the same vertical height
- Air resistance neglected
- g treated as constant over the trajectory
Do NOT use when
- Launch and landing heights differ (use full kinematics)
- Significant air drag (e.g. table-tennis ball, badminton shuttle)
- Variation of g (ballistic trajectories spanning large altitude changes)
More in Kinematics: 14 exam traps and mistakes · 1 formula · 10 question patterns from its other lessons.
Motion in a Plane questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 3, p.38
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →