Motion in a Plane

Motion in a PlaneNEET Physics · Class 11 · NCERT Chapter 3

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Motion in a Plane. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Vector addition (parallelogram + components)

Vector component resolver

Projectile motion simulator (live trajectory)

Complementary angles: same range

Horizontal projectile (off a cliff)

Centripetal acceleration calculator

Uniform circular motion visualiser

Vectors and components

Add two vectors with the parallelogram law and resolve any vector into perpendicular components.

Vectors

Vector addition (parallelogram + components)

Drag the magnitudes and the angle between two vectors to see how the resultant changes. The orange arrow is A + B.

|A|: 40

|B|: 30

Angle θ between them: 60°

Resultant

|R| = 60.83, α = 25.3°

R = A^{2} + B^{2} + 2 A B cos θ = 4 0^{2} + 3 0^{2} + 2 (40) (30) cos 60° = 60.83

tan α = \frac{B sin θ}{A + B cos θ} \Rightarrow α = 25.3°

Try this

Set θ = 0 (vectors parallel) → R = A + B (maximum).
Set θ = 180 (vectors anti-parallel) → R = |A − B| (minimum).
Set θ = 90 → R = √(A² + B²) — the Pythagorean case.
Try A = B with θ = 120° → R = A. (Asked in NEET 2014.)

Vectors

Vector component resolver

Adjust the magnitude and the angle with the x-axis to see how a vector splits into perpendicular components.

Magnitude |A|: 50

Angle θ with x-axis: 35°

A_{x} = A cos θ = 50 cos 35° = 40.96

A_{y} = A sin θ = 50 sin 35° = 28.68

Try this

θ = 0 → all of A is along x: Aₓ = A, Aᵧ = 0.
θ = 90 → all of A is along y: Aₓ = 0, Aᵧ = A.
θ = 45 → equal split: Aₓ = Aᵧ = A/√2.
NEET tip: most projectile problems start by resolving the launch velocity into u cos θ (horizontal) and u sin θ (vertical).

Projectile motion

NEET's most-tested topic in this chapter. The simulator gives you the parabola, range, max height and time of flight live, plus a side-by-side comparison of complementary launch angles.

Projectile motion

Projectile motion simulator

Drag u and θ to see the parabolic trajectory live, with range, max height and time of flight updating instantly.

Initial speed u: 30 m/s

Launch angle θ: 45°

Range

90.0 m

Max height

22.5 m

Time of flight

4.24 s

Working (g = 10 m/s²)

R = \frac{u ^{2} sin 2 θ}{g} = \frac{3 0 ^{2} sin 90°}{10} = 90.00 m

H = \frac{u ^{2} sin ^{2} θ}{2 g} = 22.50 m, T = \frac{2 u sin θ}{g} = 4.24 s

Try this

Set θ = 45° → R is maximum (= u²/g). Anything else gives smaller R.
Set θ = 30° vs θ = 60° at the same u → equal range but different heights and times of flight.
Set θ near 90° → R shrinks because the projectile spends almost all its energy going up.
Double u → R quadruples (R ∝ u²) at the same θ.

Projectile motion

Complementary angles: same range

Two projectiles launched at angles θ and (90° − θ) cover the same horizontal range. Move the slider to see the two parabolas overlap at the landing point.

Initial speed u: 40 m/s (same for both)

Lower angle θ: 30° (complement: 60°)

θ = 30°

R = 138.56 m, T = 4.00 s, H = 20.00 m

θ = 60°

R = 138.56 m, T = 6.93 s, H = 60.00 m

Both projectiles land at the same range — the complementary angles trick.

R (θ) = \frac{u ^{2} sin 2 θ}{g}, R (90° - θ) = \frac{u ^{2} sin ( 180° - 2 θ )}{g} = \frac{u ^{2} sin 2 θ}{g} .

Try this

θ = 30 vs 60 → same range. Trajectories are different but they land in the same spot.
θ = 15 vs 75 → same range, very different shapes (one flat, one tall).
The smaller angle has shorter time of flight; the larger angle reaches greater height.

Horizontal projectile

Horizontal projectile (off a cliff)

Drop an object with a horizontal launch speed u from a cliff of height h. The simulator shows the parabola, time of flight and landing speed.

Cliff height h: 80 m

Horizontal speed u: 20 m/s

Time to ground

4.00 s

Horizontal distance

80.0 m

Vertical speed at landing

40.00 m/s

Total speed at landing

44.72 m/s

T = \frac{2 h}{g} = \frac{2 \times 80}{10} = 4.00 s

R = u T = 20 \times 4.00 = 80.00 m

v = u^{2} + 2 g h = 44.72 m/s

Try this

Time to fall depends only on h, not on u — two stones thrown horizontally with different speeds land at the same time.
Doubling the cliff height multiplies the time by √2, not by 2 (because h ∝ t²).
Doubling u doubles R but does not change T.

Uniform circular motion

Centripetal acceleration is the heart of every circular motion problem in NEET. The visualiser shows why the velocity and acceleration are always perpendicular.

Circular motion

Centripetal acceleration calculator

Adjust the speed and radius for an object in uniform circular motion. The widget computes centripetal acceleration, period, angular velocity and centripetal force.

Speed v: 20 m/s

Radius r: 50 m

Mass m: 1000 kg

Centripetal acc.

8.00 m/s²

Centripetal force

8000 N

Angular vel ω

0.400 rad/s

Time period T

15.71 s

a_{c} = \frac{v ^{2}}{r} = \frac{2 0 ^{2}}{50} = 8.00 m/s^{2}

F_{c} = m a_{c} = 1000 \times 8.00 = 8000 N

Try this

Doubling the speed quadruples the centripetal acceleration (and force) at the same radius.
Doubling the radius halves the centripetal acceleration at the same speed.
NEET tip: at the highest point of vertical circular motion, gravity provides part of the centripetal force.

Circular motion

Uniform circular motion visualiser

Watch a particle move on a circle. The blue arrow is the velocity (tangent); the orange arrow is the centripetal acceleration pointing toward the centre.

Angular velocity ω: 1.50 rad/s

At every instant the velocity is tangent to the circle and the acceleration is centripetal (toward the centre). They are always perpendicular.

v = r ω, a_{c} = ω^{2} r, T = \frac{2 π}{ω} = 4.19 s

Try this

Pause and look: the velocity is always perpendicular to the radius and the centripetal acceleration is always along the radius pointing inward.
Increase ω: the particle moves faster but always on the same circle. Centripetal acceleration grows as ω².
In uniform circular motion, the speed is constant but the velocity vector keeps changing direction. That is why there is an acceleration even though |v| is fixed.

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Class 11 · Ch 2

Motion in a Straight Line

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Class 11 · Ch 4

Laws of Motion

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