PhysicsISC Class 11 · Chapter 3

Motion in a Plane ISC Class 11: Vectors and Projectile Motion

Updated 15 October 2026 · Fact-checked

Motion in a Plane is Chapter 3 of ISC Class 11 Physics. It covers scalars and vectors, unit, null and coplanar vectors, the triangle and parallelogram laws, subtraction, resolution into rectangular components, scalar and vector products, projectile motion (trajectory, time of flight, maximum height, range) and an introduction to uniform circular motion.

What Motion in a Plane is about

In two dimensions, direction matters as much as size, so we need vectors. A scalar has magnitude only (mass, time, work); a vector has magnitude and direction and obeys the laws of vector addition (displacement, velocity, force, torque).

Vector algebra. Vectors are added head to tail (triangle law) or from a common tail (parallelogram law): $R = A^{2} + B^{2} + 2 A B cos θ$ , with direction $tan α = \frac{B s i n θ}{A + B c o s θ}$ . Subtraction is adding the negative vector. A unit vector $\hat{A} = \frac{A}{A}$ gives direction; î, ĵ and k̂ point along x, y and z. Any vector can be resolved into rectangular components: $A = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ , with $A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$ . The scalar (dot) product $A \cdot B = A B cos θ$ gives a scalar (work W = F · s). The vector (cross) product $A \times B = A B sin θ \overset{n}{^}$ gives a vector perpendicular to both (torque τ = r × F).

Projectile motion. A body thrown at speed u at angle θ to the horizontal moves with constant horizontal velocity $u cos θ$ and constant downward acceleration g. Treating the two motions independently gives a parabolic path, time of flight $T = \frac{2 u s i n θ}{g}$ , maximum height $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ and range $R = \frac{u ^{2} s i n 2 θ}{g}$ , greatest at 45°. Projectiles on an inclined plane are not in the syllabus. Uniform circular motion (speed constant, direction changing) needs a centripetal acceleration $\frac{v ^{2}}{r}$ ; its dynamics is treated in Laws of Motion.

Weightage: Motion in a Plane belongs to unit 2 of the ISC Class XI Physics syllabus. CISCE weights units 1-3 together (Physical World and Measurement; Kinematics; Laws of Motion) at 23 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Scalars, vectors and kinds of vectors: A scalar is fully described by a number and a unit (mass, speed, work, energy, time, temperature). A vector needs a direction too and adds by the triangle law (displacement, velocity, acceleration, force, momentum, torque). A vector is drawn as an arrow and written $A$ (or bold A), its magnitude |A| or A. Position vector $r$ : from the origin to the point. Displacement vector: $Δ r = r_{2} - r_{1}$ . Equal vectors: same magnitude and direction (they may start anywhere). Negative vector: same magnitude, opposite direction. Null (zero) vector $0$ : zero magnitude, arbitrary direction (e.g. the displacement after a round trip). Coplanar vectors: lie in the same plane. One-, two- and three-dimensional vectors need one, two or three components. Note: a quantity with direction is not always a vector; electric current has a direction along a wire but adds as a scalar.; Worked example
A particle moves from point P (2, 3) m to point Q (5, −1) m. Write the position vectors of P and Q and find the displacement vector and its magnitude.
Try it first, then show the solutionSolution
$r_{P} = 2 \hat{i} + 3 \hat{j}$ ; $r_{Q} = 5 \hat{i} - \hat{j}$
$Δ r = r_{Q} - r_{P} = 3 \hat{i} - 4 \hat{j}$
$∣Δ r ∣ = 9 + 16 = 5 m$
Answer: $(3 \hat{i} - 4 \hat{j})$ m; 5 m
Triangle and parallelogram laws of addition: Triangle law: place the tail of $B$ at the head of $A$ ; the resultant runs from the tail of $A$ to the head of $B$ . Parallelogram law: if two vectors from a point form adjacent sides of a parallelogram, the diagonal from that point is their resultant. With angle θ between them: magnitude $R = A^{2} + B^{2} + 2 A B cos θ$ ; direction from $A$ : $tan α = \frac{B s i n θ}{A + B c o s θ}$ . Special cases: θ = 0°, R = A + B (maximum); θ = 180°, R = |A − B| (minimum); θ = 90°, $R = A^{2} + B^{2}$ . Addition is commutative and associative (proofs not required). Subtraction: $A - B = A + (- B)$ , magnitude $A^{2} + B^{2} - 2 A B cos θ$ . Multiplying by a real number n scales the magnitude by |n| and reverses the direction if n < 0.; Worked example
Two forces of 3 N and 4 N act at a point with an angle of 60° between them. Find the magnitude and direction of their resultant.
Try it first, then show the solutionSolution
$R = 9 + 16 + 2 (3) (4) cos 6 0^{\circ} = 37 = 6.08 N$
$tan α = \frac{4 s i n 6 0 ^{\circ}}{3 + 4 c o s 6 0 ^{\circ}} = \frac{3.464}{5} = 0.693$
α = 34.7° from the 3 N force
Answer: 6.08 N at 34.7° to the 3 N force
Unit vectors and resolution into components: A unit vector has magnitude 1 and shows direction only: $\hat{A} = \frac{A}{∣ A ∣}$ . î, ĵ, k̂ are the unit vectors along the x, y and z axes. Resolution splits a vector into components along chosen directions. In a plane, a vector of magnitude A at angle θ to the x-axis has rectangular components $A_{x} = A cos θ$ and $A_{y} = A sin θ$ , so $A = A_{x} \hat{i} + A_{y} \hat{j}$ , $A = A_{x}^{2} + A_{y}^{2}$ and $tan θ = \frac{A _{y}}{A _{x}}$ . In three dimensions, $A = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ and $A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$ . Adding vectors in component form is easiest: add the î parts, the ĵ parts and the k̂ parts separately.; Worked example
Find the unit vector along $A = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ , and the rectangular components of a 20 N force at 30° to the x-axis.
Try it first, then show the solutionSolution
$∣ A ∣ = 9 + 16 + 144 = 13$ , so $\hat{A} = \frac{3 i ^ + 4 j ^ - 12 k ^}{13}$
$F_{x} = 20 cos 3 0^{\circ} = 17.3 N$ ; $F_{y} = 20 sin 3 0^{\circ} = 10 N$
Answer: $\frac{1}{13} (3 \hat{i} + 4 \hat{j} - 12 \hat{k})$ ; 17.3 N and 10 N
Scalar (dot) product: $A \cdot B = A B cos θ$ , a scalar. In components, $A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ . Properties: commutative ( $A \cdot B = B \cdot A$ ); distributive; $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ ; $A \cdot A = A^{2}$ . Two non-zero vectors are perpendicular if their dot product is zero. Uses: the angle between vectors, $cos θ = \frac{A \cdot B}{A B}$ ; the component of $A$ along $B$ , $A \cdot \hat{B}$ ; work W = F · s and power P = F · v.; Worked example
A force $F = (3 \hat{i} + 4 \hat{j})$ N moves a body through $s = (2 \hat{i} + 5 \hat{j})$ m. Find the work done. Show also that $2 \hat{i} + 3 \hat{j} + \hat{k}$ and $\hat{i} - 2 \hat{j} + 4 \hat{k}$ are perpendicular.
Try it first, then show the solutionSolution
W = F · s = 3(2) + 4(5) = 26 J
Dot product = 2(1) + 3(−2) + 1(4) = 2 − 6 + 4 = 0, so cos θ = 0 and θ = 90°.
Answer: 26 J; the dot product is zero, so they are perpendicular
Vector (cross) product: $A \times B = A B sin θ \overset{n}{^}$ , where $\overset{n}{^}$ is perpendicular to the plane of $A$ and $B$ , given by the right-hand rule (curl the fingers from A to B; the thumb gives $\overset{n}{^}$ ). Properties: not commutative, $A \times B = - B \times A$ ; distributive; $A \times A = 0$ (parallel vectors give zero); $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ (cyclic) and the reverse order gives a minus sign. In components it is the determinant with rows (î, ĵ, k̂), ( $A_{x}, A_{y}, A_{z}$ ), ( $B_{x}, B_{y}, B_{z}$ ). Uses: torque τ = r × F, angular momentum L = r × p, and the area of a parallelogram $∣ A \times B ∣$ (half of it for a triangle).; Worked example
A force $F = (3 \hat{i} + 4 \hat{j})$ N acts at the point $r = (\hat{i} + 2 \hat{j})$ m. Find the torque about the origin.
Try it first, then show the solutionSolution
$τ = r \times F = (\hat{i} + 2 \hat{j}) \times (3 \hat{i} + 4 \hat{j})$
$= 4 (\hat{i} \times \hat{j}) + 6 (\hat{j} \times \hat{i}) = 4 \hat{k} - 6 \hat{k} = - 2 \hat{k}$
Answer: $- 2 \hat{k}$ N m (2 N m, clockwise seen from +z)
Projectile motion: trajectory, time of flight, height and range: A projectile launched from the ground with speed u at angle θ (air resistance ignored): horizontal motion is uniform, $x = (u cos θ) t$ ; vertical motion is uniformly accelerated, $y = (u sin θ) t - \frac{1}{2} g t^{2}$ . Eliminating t gives the trajectory $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ , a parabola. Time of flight (y = 0): $T = \frac{2 u s i n θ}{g}$ . Maximum height (vertical velocity zero): $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ . Horizontal range: $R = (u cos θ) T = \frac{u ^{2} s i n 2 θ}{g}$ , a maximum $\frac{u ^{2}}{g}$ at θ = 45°; angles θ and (90° − θ) give the same range. Velocity at any instant: $v_{x} = u cos θ$ , $v_{y} = u sin θ - g t$ , $v = v_{x}^{2} + v_{y}^{2}$ at angle $tan^{- 1} \frac{v _{y}}{v _{x}}$ to the horizontal. At the top the speed is $u cos θ$ (not zero) and velocity is perpendicular to g.; Worked example
A ball is projected at 20 m s⁻¹ at 30° above the horizontal. Find the time of flight, maximum height and range. (g = 10 m s⁻²)
Try it first, then show the solutionSolution
$T = \frac{2 ( 20 ) ( 0.5 )}{10} = 2 s$
$H = \frac{400 \times 0.25}{20} = 5 m$
$R = \frac{400 \times s i n 6 0 ^{\circ}}{10} = 34.6 m$
Answer: 2 s; 5 m; 34.6 m
Horizontal projection from a height: A body thrown horizontally at speed u from height h starts with zero vertical velocity. Time to land: $t = \frac{2 h}{g}$ (the same as for a body dropped from that height). Horizontal distance: $x = u \frac{2 h}{g}$ . Path: $y = \frac{g x ^{2}}{2 u ^{2}}$ measured downward from the launch point, a parabola. Velocity on landing: $v = u^{2} + 2 g h$ at angle $tan^{- 1} \frac{g t}{u}$ below the horizontal. Horizontal and vertical motions are independent: a ball fired horizontally and one dropped at the same instant from the same height land together.; Worked example
A ball is thrown horizontally at 15 m s⁻¹ from a cliff 45 m high. Find the time to land, the horizontal distance and the velocity on landing. (g = 10 m s⁻²)
Try it first, then show the solutionSolution
$t = \frac{2 \times 45}{10} = 3 s$ ; x = 15 × 3 = 45 m
$v_{y} = 10 \times 3 = 30 m s^{- 1}$ ; $v = 1 5^{2} + 3 0^{2} = 33.5 m s^{- 1}$
Angle below horizontal = $tan^{- 1} (30/15) = 63. 4^{\circ}$
Answer: 3 s; 45 m; 33.5 m s⁻¹ at 63.4° below the horizontal
Uniform circular motion (kinematics): In uniform circular motion a body moves on a circle of radius r at constant speed v. Its velocity is along the tangent and keeps changing direction, so the body accelerates. Angular displacement θ (radian), angular velocity $ω = \frac{d θ}{d t}$ , with $v = ω r$ , period $T = \frac{2 π}{ω} = \frac{2 π r}{v}$ and frequency $ν = \frac{1}{T}$ , so $ω = 2 π ν$ . The centripetal acceleration is directed towards the centre with magnitude $a_{c} = \frac{v ^{2}}{r} = ω^{2} r$ . Speed is constant, velocity is not; the acceleration is constant in magnitude but not in direction, so the equations of uniform acceleration do not apply. The force that produces this acceleration, banking and vertical circles are treated in Laws of Motion.; Worked example
A stone tied to a string is whirled in a horizontal circle of radius 0.5 m, making 2 revolutions per second. Find its angular speed, linear speed and centripetal acceleration.
Try it first, then show the solutionSolution
ω = 2πν = 2π × 2 = 12.6 rad s⁻¹
v = ωr = 12.6 × 0.5 = 6.28 m s⁻¹
$a_{c} = ω^{2} r = (12.57)^{2} \times 0.5 = 79.0 m s^{- 2}$ , towards the centre
Answer: 12.6 rad s⁻¹; 6.28 m s⁻¹; 79 m s⁻² (about 8g)

Physics in real life: Motion in a Plane

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Adding velocities: crossing a river

Suppose you row a boat at 4 m s⁻¹ straight across a river 80 m wide while the current flows at 3 m s⁻¹.

Find the boat's resultant velocity, the time to cross and how far downstream it lands.

Try it first, then show the solutionSolution

$v = 4^{2} + 3^{2} = 5 m s^{- 1}$ at $tan^{- 1} (3/4) = 36. 9^{\circ}$ to the line straight across
Time = 80/4 = 20 s (only the across component matters); drift = 3 × 20 = 60 m

Answer: 5 m s⁻¹ at 36.9° downstream; 20 s; it lands 60 m downstream.

Memory hook: The current changes where you land, not when.

Maximum range of a throw

Imagine a fielder who can throw a cricket ball at 25 m s⁻¹ and wants it to travel as far as possible before it first bounces.

What is the greatest distance he can throw it, and at what angle? (g = 10 m s⁻², air resistance ignored)

Try it first, then show the solutionSolution

Maximum range at θ = 45°: $R = \frac{u ^{2}}{g} = \frac{625}{10}$
R = 62.5 m

Answer: 62.5 m at 45°; real throws use a lower angle because air resistance shortens the path.

Memory hook: 45° for the farthest flight.

Work as a dot product

Suppose you pull a wheeled suitcase 20 m along a level platform with a force of 50 N along the handle, which makes 60° with the ground.

How much work do you do on the suitcase, and what part of your force does it?

Try it first, then show the solutionSolution

Only the component along the motion does work: $50 cos 6 0^{\circ} = 25 N$
W = F · s = 50 × 20 × cos 60° = 500 J

Answer: 500 J; the horizontal 25 N component does the work, while the vertical component only lifts some weight off the wheels.

Memory hook: Work counts only the part of the force along the motion.

Topics in this chapter

Motion in a Plane formula sheet

Parallelogram law: $R = A^{2} + B^{2} + 2 A B cos θ, tan α = \frac{B s i n θ}{A + B c o s θ}$; α measured from $A$ .
Magnitude and unit vector: $A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}, \hat{A} = \frac{A}{A}$; Components: $A_{x} = A cos θ$ , $A_{y} = A sin θ$ .
Dot product: $A \cdot B = A B cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$; Zero for perpendicular vectors; W = F · s.
Cross product: $∣ A \times B ∣ = A B sin θ, \hat{i} \times \hat{j} = \hat{k}$; Zero for parallel vectors; τ = r × F.
Trajectory: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$; A parabola.
Time of flight: $T = \frac{2 u s i n θ}{g}$; Launch and landing at the same level.
Maximum height: $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$; Speed at the top = $u cos θ$ .
Horizontal range: $R = \frac{u ^{2} s i n 2 θ}{g}, R_{ma x} = \frac{u ^{2}}{g} (θ = 4 5^{\circ})$; θ and 90° − θ give the same range.
Horizontal projection: $t = \frac{2 h}{g}, x = u \frac{2 h}{g}$; Landing speed $u^{2} + 2 g h$ .
Uniform circular motion: $v = ω r, a_{c} = \frac{v ^{2}}{r} = ω^{2} r, ω = 2 π ν$; Acceleration towards the centre.

Motion in a Plane: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Parallelogram law of vector addition

Deriving the magnitude and direction of the resultant.

Draw

Vector $A$ as OP along the horizontal and $B$ as OQ at angle θ to it, from the same point O.
Complete the parallelogram OPSQ; draw the diagonal OS as the resultant $R$ .
Extend OP to N and drop the perpendicular SN; mark angle θ at P between PN and PS, and α between OS and OP.

Label

Watch out: In triangle OSN: OS² = (OP + PN)² + SN², with PN = B cos θ and SN = B sin θ; this gives both R and tan α.

Path of a projectile

Deriving the trajectory, T, H and R, or showing the velocity components at points on the path.

Draw

x and y axes with the launch point at O; a symmetric parabola landing at distance R.
At O, the velocity u at angle θ split into $u cos θ$ (horizontal) and $u sin θ$ (vertical).
At the top, only the horizontal velocity $u cos θ$ ; at a later point, $u cos θ$ and a downward $v_{y}$ ; the acceleration g shown downward everywhere.

Label

O
u
θ
$u cos θ$
$u sin θ$
H (maximum height)
R (range)
g

Watch out: Keep the horizontal arrow the same length at every point: horizontal velocity never changes.

How the school exam asks Motion in a Plane

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Vector definitions	Distinguish scalar and vector; define unit, null, equal and coplanar vectors; give examples.	One line each with an example; for null vectors give the round-trip displacement.
Resultant and resolution	Find the resultant of two vectors at an angle; find components; find a unit vector.	Use the parallelogram formula or components; give both magnitude and direction.
Dot and cross products	Angle between vectors, perpendicularity condition, work, torque, area of a parallelogram.	Write the component formula; for cross products use the determinant and the cyclic rule î × ĵ = k̂.
Projectile derivations	Show the path is a parabola; derive T, H and R; show maximum range at 45°.	Separate horizontal (uniform) and vertical (accelerated) motion, then eliminate t.
Projectile numericals	Ground-to-ground projectiles, horizontal throws from a cliff or aircraft, complementary angles.	Resolve u first; use the vertical motion to get time, then the horizontal motion for distance.
Uniform circular motion	Find ω, v and centripetal acceleration; explain why a body at constant speed accelerates.	Convert rpm or rev s⁻¹ to rad s⁻¹ (multiply by 2π); state the direction (towards the centre).

Tricks to remember Motion in a Plane

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cyclic i-j-k
Write i → j → k → i in a circle: going clockwise gives +, against gives − (î × ĵ = k̂, ĵ × î = −k̂).
Use it when: Cross products in component form.
Dot = cos, cross = sin
Dot uses cos (zero when perpendicular); cross uses sin (zero when parallel).
Use it when: Perpendicular and parallel conditions.
THR of a projectile
T has sin θ, H has sin² θ, R has sin 2θ; all have u, and T has u, H and R have u².
Use it when: Recalling the three projectile formulae.
Complementary twins
15° and 75°, 30° and 60°: same range, different heights and times.
Use it when: Two-angle range problems.
Horizontal is boring
Horizontal velocity never changes; all the action (and g) is vertical.
Use it when: Setting up every projectile problem.

Motion in a Plane: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Equal sum and difference · 1 mark · MCQExam level

If $∣ A + B ∣ = ∣ A - B ∣$ , the angle between $A$ and $B$ is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°

Try it first, then show the solutionSolution

Squaring: A² + B² + 2AB cos θ = A² + B² − 2AB cos θ, so cos θ = 0 and θ = 90°.

Answer: (c)

Example 2: Velocity and acceleration at the top · 1 mark · Assertion-reasonExam level

Assertion: At the highest point of its path, the velocity of a projectile is perpendicular to its acceleration.
Reason: At the highest point the vertical component of velocity is zero, while the acceleration is vertically downward.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

At the top only $u cos θ$ (horizontal) remains, and g is vertical, so they are at 90°: the Reason explains the Assertion.

Answer: (a)

Example 3: Angle between two vectors · 2 marks · Short answerHarder exam level

Find the angle between $A = 2 \hat{i} + 2 \hat{j} - \hat{k}$ and $B = 6 \hat{i} - 3 \hat{j} + 2 \hat{k}$ . [2]

Try it first, then show the solutionSolution

$A \cdot B = 12 - 6 - 2 = 4$ ; A = √9 = 3; B = √49 = 7
cos θ = 4/21 = 0.190, so θ = 79.0°

Answer: About 79°

Example 4: Two angles for the same range · 3 marks · Short answerHarder exam level

Show that the horizontal range is the same for angles of projection θ and (90° − θ). A ball thrown at 20 m s⁻¹ must land 20 m away; find the two possible angles. (g = 10 m s⁻²) [3]

Try it first, then show the solutionSolution

$R (9 0^{\circ} - θ) = \frac{u ^{2} s i n ( 18 0 ^{\circ} - 2 θ )}{g} = \frac{u ^{2} s i n 2 θ}{g} = R (θ)$
$sin 2 θ = \frac{R g}{u ^{2}} = \frac{200}{400} = 0.5$ , so 2θ = 30° or 150°.
θ = 15° or 75°

Answer: 15° or 75°

Example 5: Area from a cross product · 3 marks · Short answerHarder exam level

Two sides of a triangle are $A = 3 \hat{i} + 4 \hat{j}$ and $B = - 3 \hat{i} + 7 \hat{j}$ . Find the area of the triangle and a unit vector perpendicular to its plane. [3]

Try it first, then show the solutionSolution

$A \times B = (3 \times 7 - 4 \times (- 3)) \hat{k} = 33 \hat{k}$
Area of triangle = ½ |A × B| = 16.5 square units
Unit vector perpendicular = $\frac{33 k ^}{33} = \hat{k}$ (or −k̂)

Answer: 16.5 square units; ±k̂

Example 6: Projectile from the ground · 5 marks · Long answerHarder exam level

A body is projected with speed u at angle θ to the horizontal.
(a) Show that its path is a parabola. [2]
(b) Derive expressions for its time of flight and maximum height. [2]
(c) Find the maximum height if u = 30 m s⁻¹ and θ = 60°. (g = 10 m s⁻²) [1]

Try it first, then show the solutionSolution

(a) $x = (u cos θ) t$ so $t = \frac{x}{u c o s θ}$ ; $y = (u sin θ) t - \frac{1}{2} g t^{2}$
Substituting: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ , of the form $y = a x - b x^{2}$ : a parabola.
(b) At landing y = 0: $t (u sin θ - \frac{1}{2} g t) = 0$ , so $T = \frac{2 u s i n θ}{g}$ .
At the top $v_{y} = 0$ : $0 = u^{2} sin^{2} θ - 2 g H$ , so $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ .
(c) $H = \frac{900 \times 0.75}{20} = 33.75 m$

Answer: (c) 33.75 m

Example 7: Range and height compared · 3 marks · Short answerHOTS / competency

Find the angle of projection for which the horizontal range equals (a) four times the maximum height, and (b) the maximum height. [3]

Try it first, then show the solutionSolution

$\frac{R}{H} = \frac{2 u ^{2} s i n θ c o s θ / g}{u ^{2} s i n ^{2} θ /2 g} = \frac{4}{t a n θ}$
(a) R = 4H: tan θ = 1, θ = 45°.
(b) R = H: tan θ = 4, θ = 76.0°.

Answer: (a) 45° (b) about 76°

Example 8: Dropping a relief packet · 3 marks · Short answerHOTS / competency

An aircraft flies horizontally at 720 km h⁻¹ at a height of 2 km and releases a packet. Where does the packet land relative to the point of release, and where is the aircraft at that moment? (g = 10 m s⁻², ignore air resistance) [3]

Try it first, then show the solutionSolution

u = 720 × 5/18 = 200 m s⁻¹ horizontally; fall time $t = \frac{2 \times 2000}{10} = 20 s$
Horizontal distance = 200 × 20 = 4000 m = 4 km ahead of the release point.
The packet keeps the aircraft's horizontal velocity, so the aircraft is directly above it when it lands.

Answer: 4 km ahead of the release point; the aircraft is vertically above it

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

A free kick over the wall

Suppose a footballer kicks a ball from the ground at 20 m s⁻¹ at 37° above the horizontal. A wall of players 6 m high stands 20 m away. Take sin 37° = 0.6, cos 37° = 0.8 and g = 10 m s⁻², and ignore air resistance.

(i) · 1 mark
Find the time of flight.
Answer
Answer: 2.4 s
T = 2 × 20 × 0.6/10 = 2.4 s.
(ii) · 1 mark
Find the maximum height reached.
Answer
Answer: 7.2 m
H = 400 × 0.36/20 = 7.2 m.
(iii) · 1 mark
Find the horizontal range.
Answer
Answer: 38.4 m
R = 400 × 2 × 0.6 × 0.8/10 = 38.4 m.
(iv) · 2 marks
Does the ball clear the wall? Show your working.
Answer
Answer: Yes, at about 7.19 m it clears the 6 m wall
Horizontal speed 16 m s⁻¹, so it reaches the wall at t = 20/16 = 1.25 s; y = 12(1.25) − 5(1.25)² = 15 − 7.81 = 7.19 m > 6 m.

Two ropes on a hook

Suppose two ropes pull on a hook fixed in a wall: one with 30 N towards the east (x-axis) and the other with 40 N towards the north (y-axis).

(i) · 1 mark
Write the resultant force in unit-vector form and find its magnitude.
Answer
Answer: $(30 \hat{i} + 40 \hat{j})$ N; 50 N
$R = 3 0^{2} + 4 0^{2} = 50 N$ .
(ii) · 1 mark
Find the direction of the resultant.
Answer
Answer: 53.1° north of east
tan α = 40/30, α = 53.1°.
(iii) · 1 mark
Find the unit vector along the resultant.
Answer
Answer: $0.6 \hat{i} + 0.8 \hat{j}$
$\frac{30 i ^ + 40 j ^}{50}$ .
(iv) · 2 marks
If the hook pulls loose and moves through $(2 \hat{i} + \hat{j})$ m under these forces, find the work done.
Answer
Answer: 100 J
W = F · s = 30(2) + 40(1) = 100 J.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Two forces of 5 N and 12 N act at right angles. Find the magnitude and direction of their resultant. (2 marks)

Step	Marks
$R = 5^{2} + 1 2^{2} = 13 N$	1
tan α = 12/5, α = 67.4° to the 5 N force	1

A ball is thrown horizontally at 10 m s⁻¹ from the top of a building 20 m high. Find the time to reach the ground, the horizontal distance and the speed on landing. (g = 10 m s⁻²) (3 marks)

Step	Marks
$t = 2 h / g = 4 = 2 s$	1
x = 10 × 2 = 20 m	1
$v_{y} = 20 m s^{- 1}$ ; $v = 100 + 400 = 22.4 m s^{- 1}$	1

Derive an expression for the horizontal range of a projectile and show that it is greatest at 45°. Find the maximum range for u = 25 m s⁻¹ (g = 10 m s⁻²). (5 marks)

Step	Marks
Resolve: horizontal $u cos θ$ (constant), vertical $u sin θ$ with acceleration −g	1
Time of flight $T = \frac{2 u s i n θ}{g}$	1
$R = u cos θ \times T = \frac{u ^{2} s i n 2 θ}{g}$	1
R is greatest when sin 2θ = 1, i.e. θ = 45°, giving $R_{ma x} = u^{2} / g$	1
$R_{ma x} = 625/10 = 62.5 m$	1

Where students lose marks in Motion in a Plane

The slips examiners see most often, and what to write instead.

Mistake: Adding magnitudes instead of vectors
Students add 3 N and 4 N to get 7 N whatever the angle between them.
Wrong working
3 N and 4 N at 60°: R = 7 N.
Right working
$R = 9 + 16 + 24 cos 6 0^{\circ} = 6.08 N$ .
Do this: Use the parallelogram formula or add components.
Usually costs: 1-2 marks.
Mistake: Wrong sign in a cross product
Students treat the cross product as commutative or forget that ĵ × î = −k̂.
Wrong working
$(\hat{i} + 2 \hat{j}) \times (3 \hat{i} + 4 \hat{j}) = 4 \hat{k} + 6 \hat{k} = 10 \hat{k}$ .
Right working
$4 (\hat{i} \times \hat{j}) + 6 (\hat{j} \times \hat{i}) = 4 \hat{k} - 6 \hat{k} = - 2 \hat{k}$ .
Do this: Use the determinant, or the cyclic rule with signs.
Usually costs: 1 mark.
Mistake: Saying the velocity at the top of a projectile's path is zero
Students carry over the vertical-throw result to an oblique projectile.
Wrong working
At the highest point, v = 0 and the kinetic energy is zero.
Right working
At the top only $v_{y} = 0$ ; the speed is $u cos θ$ and the kinetic energy is $\frac{1}{2} m u^{2} cos^{2} θ$ .
Do this: Remember that the horizontal velocity never changes.
Usually costs: 1 mark.
Mistake: Using the ground-to-ground formulae for a throw from a height
T = 2u sin θ/g and R = u² sin 2θ/g assume the projectile lands at its launch level.
Wrong working
Thrown horizontally from a 20 m building: θ = 0, so R = 0.
Right working
Use $t = 2 h / g$ from the vertical motion, then x = ut.
Do this: When launch and landing levels differ, work from the two component equations.
Usually costs: 2-3 marks.
Mistake: Mixing degrees and rev s⁻¹ in circular motion
Students put the frequency straight into v = ωr without the factor 2π.
Wrong working
2 rev s⁻¹, r = 0.5 m: v = 2 × 0.5 = 1 m s⁻¹.
Right working
ω = 2π × 2 = 12.6 rad s⁻¹; v = 6.28 m s⁻¹.
Do this: Convert to rad s⁻¹ first: ω = 2πν.
Usually costs: 1 mark.

Practice questions on Motion in a Plane

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of the following is a vector quantity?
(a) work
(b) pressure
(c) torque
(d) energy
Hint
Which one is a cross product?
Answer
(c) torque (τ = r × F)
Q2 · 1 mark · MCQExam level
$\hat{k} \times \hat{j}$ equals:
(a) î
(b) −î
(c) 0
(d) k̂
Hint
Against the cyclic order.
Answer
(b) −î
Q3 · 1 mark · Very short answerExam level
What is a null vector? Give one example.
Hint
Zero magnitude.
Answer
A vector of zero magnitude and arbitrary direction, e.g. the displacement of a body that returns to its starting point.
Q4 · 1 mark · Very short answerExam level
What is the angle between the velocity and acceleration of a body in uniform circular motion?
Hint
Tangent and radius.
Answer
90°: velocity is along the tangent, acceleration towards the centre.
Q5 · 2 marks · Short answerExam level
Find the rectangular components of a 10 N force acting at 60° to the x-axis in the xy-plane.
Hint
cos and sin.
Answer
$F_{x} = 10 cos 6 0^{\circ} = 5 N$ ; $F_{y} = 10 sin 6 0^{\circ} = 8.66 N$ .
Q6 · 2 marks · Short answerHarder exam level
If $A = 3 \hat{i} + 4 \hat{j}$ and $B = \hat{i} - 2 \hat{j}$ , find $A + B$ , $A - B$ and $∣ A - B ∣$ .
Hint
Component by component.
Answer
$4 \hat{i} + 2 \hat{j}$ ; $2 \hat{i} + 6 \hat{j}$ ; $40 = 6.32$ .
Q7 · 2 marks · Short answerHarder exam level
For what value of m are $A = 2 \hat{i} + m \hat{j} + \hat{k}$ and $B = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$ perpendicular?
Hint
Dot product zero.
Answer
8 − 2m − 2 = 0, so m = 3.
Q8 · 2 marks · Short answerHarder exam level
A projectile launched from the ground lands 60 m away after 3 s. Find the speed and angle of projection and the maximum height. (g = 10 m s⁻²)
Hint
$u_{x} = R / T$ ; $u_{y} = g T /2$ .
Answer
$u_{x} = 20$ , $u_{y} = 15$ m s⁻¹: u = 25 m s⁻¹ at 36.9°; H = 15²/20 = 11.25 m.
Q9 · 2 marks · Short answerHarder exam level
Two forces of equal magnitude F have a resultant also of magnitude F. Find the angle between them.
Hint
F² = 2F² + 2F² cos θ.
Answer
cos θ = −½, so θ = 120°.
Q10 · 3 marks · Short answerHarder exam level
A stone is thrown horizontally at 10 m s⁻¹ from a cliff 20 m high. Find where and with what velocity it lands. (g = 10 m s⁻²)
Hint
Vertical motion gives the time.
Answer
t = 2 s; 20 m from the foot of the cliff; $v = 1 0^{2} + 2 0^{2} = 22.4$ m s⁻¹ at $tan^{- 1} 2 = 63. 4^{\circ}$ below the horizontal.
Q11 · 3 marks · Short answerHarder exam level
A force $F = (\hat{i} + 3 \hat{j} - \hat{k})$ N acts at the point $r = (2 \hat{i} - \hat{j} + \hat{k})$ m. Find the torque about the origin and its magnitude.
Hint
Determinant.
Answer
$τ = \hat{i} (1 - 3) - \hat{j} (- 2 - 1) + \hat{k} (6 + 1) = - 2 \hat{i} + 3 \hat{j} + 7 \hat{k}$ N m; $∣ τ ∣ = 62 = 7.87$ N m.
Q12 · 5 marks · Long answerHarder exam level
(a) State the parallelogram law of vector addition and derive the magnitude of the resultant. [2]
(b) When is the resultant a maximum and when a minimum? [1]
(c) Find the resultant of forces of 5 N and 12 N acting at right angles. [2]
Hint
Drop a perpendicular from the far corner.
Answer
(a) If two vectors from a point are adjacent sides of a parallelogram, the diagonal from that point is the resultant; from the right triangle, $R^{2} = (A + B cos θ)^{2} + (B sin θ)^{2} = A^{2} + B^{2} + 2 A B cos θ$ . (b) Maximum A + B at θ = 0°; minimum |A − B| at θ = 180°. (c) R = 13 N at 67.4° to the 5 N force.
Q13 · 3 marks · Short answerHOTS / competency
For a given speed of projection, show that the maximum heights $H_{1}$ and $H_{2}$ for complementary angles satisfy $H_{1} + H_{2} = \frac{u ^{2}}{2 g}$ and $R = 4 H_{1} H_{2}$ .
Hint
Use sin θ and cos θ.
Answer
$H_{1} = \frac{u ^{2} s i n ^{2} θ}{2 g}$ , $H_{2} = \frac{u ^{2} c o s ^{2} θ}{2 g}$ , so $H_{1} + H_{2} = \frac{u ^{2}}{2 g}$ . $H_{1} H_{2} = \frac{u ^{2} s i n θ c o s θ}{2 g}$ , so $4 H_{1} H_{2} = \frac{2 u ^{2} s i n θ c o s θ}{g} = R$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for the resultant of two vectors at angle θ?Show answer
$R = A^{2} + B^{2} + 2 A B cos θ$ .
2.Condition for two vectors to be perpendicular?Show answer
Their dot product is zero.
3.What is $A \times A$ ?Show answer
The null vector.
4.Time of flight of a projectile?Show answer
$T = \frac{2 u s i n θ}{g}$ .
5.Angle for maximum range?Show answer
45°.
6.Speed at the top of a projectile's path?Show answer
$u cos θ$ .
7.Centripetal acceleration in terms of ω?Show answer
$ω^{2} r$ , towards the centre.
8.Is projectile motion on an inclined plane in the ISC syllabus?Show answer
No, it is excluded.

Best order to study it

Learn scalars, vectors and the kinds of vectors with examples.
Learn the triangle and parallelogram laws and derive the resultant formula; practise subtraction.
Practise unit vectors and resolution into rectangular components.
Learn dot and cross products with their properties and uses (work, torque, area).
Derive the trajectory, T, H and R of a projectile; practise ground and cliff problems.
Learn the kinematics of uniform circular motion (ω, v = ωr, a = v²/r).

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you give three scalars and three vectors?
Can you derive the parallelogram-law resultant and its direction?
Can you find a unit vector and the components of a vector?
Can you find the angle between two vectors with a dot product?
Can you find a torque or an area with a cross product?
Can you derive the equation of the trajectory and T, H, R?
Can you solve a horizontal-projection problem from a cliff?
Can you find the centripetal acceleration from the frequency?

Motion in a Plane: frequently asked questions

Do I need to prove that vector addition is commutative and associative?

No. The ISC syllabus says the proofs of the commutative and associative properties are not required; you should know and use the properties.

Is projectile motion on an inclined plane in the ISC Class 11 syllabus?

No. The syllabus explicitly excludes projectile motion on an inclined plane. You need the trajectory, time of flight, maximum height, horizontal range and instantaneous velocity.

Where is uniform circular motion treated in detail?

The syllabus places the details of uniform circular motion in the Laws of Motion unit: centripetal force, banking of roads and motion in a vertical circle. This chapter gives the kinematics: ω, v = ωr and a = v²/r.

Why is current not a vector although it has a direction?

Currents meeting at a junction add arithmetically, not by the triangle law. A vector must have magnitude and direction and also obey the laws of vector addition.

Why does a body in uniform circular motion accelerate if its speed is constant?

Acceleration is the rate of change of velocity, and velocity includes direction. The direction changes continuously, so there is an acceleration (towards the centre) even though the speed does not change.

Do I need the determinant method for cross products?

It is the quickest reliable method for vectors with three components. For vectors in a plane you can also expand term by term using î × ĵ = k̂ and ĵ × î = −k̂.

Based on: CISCE ISC 2028 Physics syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)