motion in a plane parabolic trajectory cannon ball x and y axis


This article is formulated according to the 4th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.

When a body moves in a plane (a two-dimensional motion) or in a space (a three-dimensional motion) then the position, displacement, velocity and acceleration of the body have two or three components respectively. Then we need to use Vectors to describe the concept of position, displacement, velocity and acceleration.

Scalar quantity

A physical quantity having only magnitude is called a scalar quantity. It is specified completely by a single number along with proper unit. Ex: mass, length, temperature, speed, charge, area etc. Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers. They follow the rules of algebra.

Vector quantity

A Physical quantity having both magnitude and direction and obey the triangle law of addition is called Vector quantity. It is represented by a number with an appropriate unit and direction. Ex: Displacement, velocity, acceleration, force, momentum etc.

Differences between Scalar quantity and Vector quantity

Scalar QuantityVector Quantity
It has only magnitudeIt has both magnitude and direction
They follow the rules of ordinary algebraThey follow the rules of vector algebra
These change when magnitude changesThese changes when magnitude changes or direction changes or both of them changes.
Ex: Mass, Length, Temperature, AreaEx: Displacement, velocity, Acceleration, Force

Representation of a vector

To represent a vector we use a bold face letters or an arrow placed over a letter.


Here O is called the initial point and P is called the terminal point. The length of the line segment OP represents the magnitude and the arrow at the endpoint indicates the direction. The magnitude of a vector is often called the absolute value and indicated by,

Classification of vectors

Parallel vectors

Two or more vectors having same direction are called parallel vectors.

Anti-parallel vectors (opposite vectors)

Vectors having opposite directions are called anti-parallel vectors (opposite vectors).

Equality of vector (Equal vectors)

Two (or more) vectors having same magnitude and direction, representing the same physical quantity are called Equal vectors.

Negative of a vector

A vector having same magnitude but having opposite direction to that of the given vector is called negative of a given vector.

Zero(Null) vector

A vector whose magnitude is zero is called Zero vector. It is represented by 0βƒ—βƒ— and the direction is not specified.

Properties of Zero vector are, 𝐴⃗ βˆ’ 𝐴⃗ = 0βƒ—βƒ—

|0βƒ—βƒ—| = 0

𝐴⃗ βˆ’ 0βƒ—βƒ— = 𝐴⃗

0βƒ—βƒ— 𝐴⃗ = 𝐴⃗ 0βƒ—βƒ— = 0βƒ—βƒ—

πœ† 0βƒ—βƒ— = 0βƒ—βƒ—

Unit vector

A vector having unit magnitude is called unit vector. Its purpose is to specify a direction. Unit vector has no dimensions and unit. If π‘Žβƒ— is a vector, then the unit vector in direction of π‘Žβƒ— is written as π‘ŽΜ‚ (read as β€œa cap”) π‘Žβƒ— = |π‘Žβƒ—| π‘ŽΜ‚, Then, mathematically unit vector can be represented as, π‘ŽΜ‚ = π‘Žβƒ— |π‘Žβƒ—|

Note: The unit vectors in the positive directions of x, y and z axes are labeled as 𝑖,Μ‚ 𝑗̂ π‘Žπ‘›π‘‘ π‘˜Μ‚ respectively.

Addition of vectors – Graphical method

Two vectors representing the same quantity in the same unit are added using the following rules.

(i) Triangle method of vector addition: Law of the triangle of vectors or Triangular law of vector addition:

If two vectors π‘Žβƒ— and 𝑏⃗⃗ are represented by two sides of a triangle in head-to-tail form, then the closing side of the triangle taken from the tail of the first to head of the second represents the vector sum of π‘Žβƒ— and 𝑏⃗.

Explanation: Consider two vectors, π‘Žβƒ— = 𝐴𝐡⃗ and 𝑏⃗⃗ = 𝐡𝐢⃗ are of same nature. The triangle ABC is completed by joining A and C. According to the triangle law of addition, 𝐴𝐢⃗ = π‘Ÿβƒ— represents the sum of π‘Žβƒ— and 𝑏⃗. π‘Žβƒ— + 𝑏⃗ = π‘Ÿβƒ— or 𝐴𝐡⃗ + 𝐡𝐢⃗ = 𝐴𝐢⃗

Note: In this procedure of vector addition, vectors are arranged head-to-tail. Hence it is called the head-to-tail method.

Properties vector addition

(a) Vector addition is commutative.

(b) Vector addition is Associative.

π‘Žβƒ— + 𝑏⃗ = π‘Ÿβƒ— = 𝑏⃗ + π‘Žβƒ— (π‘Žβƒ— + 𝑏) + 𝑐⃗ = π‘Žβƒ— + (𝑏⃗ + 𝑐⃗)

(ii) Parallelogram method of vector addition

Law of parallelogram of vectors or Parallelogram law of vector addition

If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal drawn from the common initial point represents their vector sum.

Explanation: Vector π‘Žβƒ— and 𝑏⃗ are drawn with a common initial point and parallelogram is constructed using these two vectors as two adjacent sides of a parallelogram. The diagonal originating from the common initial point is vector sum of π‘Žβƒ— and 𝑏⃗.

Subtraction vectors – Graphical method

Subtraction of vectors can be defined in terms of addition of vectors. Consider two vectors π‘Žβƒ— and 𝑏⃗ of same nature and another vector βˆ’π‘βƒ— which is opposite (negative) vector of 𝑏⃗ , then π‘Žβƒ— + ( βˆ’π‘βƒ—) = π‘Žβƒ— + 𝑏⃗

Note: Subtraction vector is neither commutative nor associative.

Multiplication of a vector by real (Scalar) number OR Scalar multiplication of a Vector

The product of a vector 𝑣⃗ and a positive number (Scalar) πœ† gives a vector, whose magnitude is changed by a factor πœ† but direction is same as that of 𝑣⃗. |πœ†π‘£βƒ—| = πœ†|𝑣⃗ | (𝑖𝑓 πœ† > 0) If πœ† is negative, the direction of the vector πœ†π‘£βƒ— is opposite to the direction of the vector 𝑣⃗ and magnitude is – πœ† times |𝑣⃗ |. If the multiplying factor πœ† is dimensionless then πœ†π‘£βƒ— have the same dimensions as that of 𝑣⃗ and is product of dimensions if πœ† has dimensions.

Resolution of vectors

Splitting a given vector into a number of components is called resolution of vectors OR The process of finding the components of a given vector is called resolution the vector.

Expressions for X and Y components of a Vector

Consider a vector 𝑂𝐴⃗ = π‘Žβƒ— in X-Y plane, which makes an angle πœƒ with the positive X-axis. Draw AM and AN perpendicular to X and Y axes respectively. Let 𝑂𝑀⃗ = π‘Žβƒ—π‘₯ and 𝑂𝑁⃗ = π‘Žβƒ—π‘¦. From parallelogram law of addition, we have 𝑂𝑀⃗ + 𝑂𝑁⃗ = 𝑂𝐴⃗⃗

π‘Žβƒ—π‘₯ + π‘Žβƒ—π‘¦ = π‘Žβƒ— (Here π‘Žβƒ—π‘₯ is x – component of π‘Žβƒ— and π‘Žβƒ—π‘¦ is y – component of π‘Žβƒ—) From βˆ†π‘‚π΄π‘€, π‘π‘œπ‘  πœƒ = 𝑂𝑀/𝑂𝐴 = π‘Žπ‘₯/π‘Ž

𝒂𝒙 = 𝒂 𝐜𝐨𝐬 𝜽

and 𝑠𝑖𝑛 πœƒ = 𝐴𝑀/𝑂𝐴 = π‘Žπ‘¦/π‘Ž

π’‚π’š = 𝒂 𝐬𝐒𝐧 𝜽

Vectoricaly 𝒂⃗𝒙 = 𝒂⃗ 𝐜𝐨𝐬 𝜽 and π’‚βƒ—π’š = 𝒂⃗𝐬𝐒𝐧 𝜽

Note: π‘Žβƒ—π‘₯ and π‘Žβƒ—π‘¦ being perpendicular are called rectangular components of π‘Žβƒ—.


Magnitude of π‘Žβƒ— is given by |π‘Žβƒ—| = π‘Ž

Now, π‘Žπ‘₯2 = π‘Ž2cos2πœƒ and π‘Žπ‘¦2 = π‘Ž2sin2πœƒ

Taking π‘Žπ‘₯2 + π‘Žπ‘¦2 = π‘Ž2cos2πœƒ + π‘Ž2sin2πœƒ

π‘Žπ‘₯2 + π‘Žπ‘¦2 = π‘Ž2(cos2πœƒ + sin2πœƒ)

π‘Žπ‘₯2 + π‘Žπ‘¦2 = π‘Ž2

𝒂 = √(π’‚π’™πŸ + π’‚π’šπŸ)


By taking π‘Žπ‘¦/π‘Žπ‘₯ = π‘Žsinπœƒ/π‘Žcosπœƒ

π‘Žπ‘¦/π‘Žπ‘₯ = tanπœƒ

𝜽 = π­πšπ§βˆ’πŸ(π’‚π’š/𝒂𝒙)

Note: (i) In terms of unit vectors, π‘Žβƒ— = π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚= π‘Žcosπœƒπ‘–Μ‚ + sinπœƒπ‘—Μ‚

where π‘Žβƒ—π‘₯ = π‘Ž cos πœƒ 𝑖̂

and π‘Žβƒ—π‘¦ = π‘Ž sin πœƒ 𝑗̂

(ii) If π‘Žβƒ— is in XYZ plane and makes an angle 𝛼, 𝛽 and 𝛾 with X, Y and Z axes respectively, then π‘Žπ‘₯ = π‘Ž cos 𝛼, π‘Žπ‘¦ = π‘Ž cos𝛽, π‘Žπ‘§ = π‘Ž cos 𝛾

and π‘Žβƒ— = π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚+ π‘Žπ‘§π‘˜Μ‚

The magnitude of π‘Žβƒ— is, π‘Ž = √(π‘Žπ‘₯2 + π‘Žπ‘¦2 + π‘Žπ‘§2).

Find the magnitude and direction of the resultant of two vectors 𝑨⃗ and 𝑩⃗ in terms of their magnitudes and angle 𝜽 between them.

Let 𝑂𝑃⃗ and 𝑂𝑄⃗ represent the two vectors 𝐴⃗ and 𝐡⃗ making an angle πœƒ. Then using the parallelogram method of vector addition 𝑂𝑆 βƒ— represents the resultant vector 𝑅⃗.

𝑅⃗ = 𝐴⃗ + 𝐡⃗

Draw SN is normal to OP extended. In βˆ†π‘†π‘ƒπ‘, π‘π‘œπ‘  πœƒ = 𝑃𝑁/𝑆𝑃

𝑃𝑁 = 𝑆𝑃 cos πœƒ = 𝐡 cos πœƒ and 𝑠𝑖𝑛 πœƒ = 𝑆𝑁/𝑆𝑃

𝑆𝑁 = 𝑆𝑃 sin πœƒ = 𝐡 sin πœƒ


From geometry, 𝑂𝑆2 = 𝑂𝑁2 + 𝑆𝑁2

𝑂𝑆2 = (𝑂𝑃 + 𝑃𝑁)2 + 𝑆𝑁2

𝑅2 = (𝐴 + 𝐡 cos πœƒ)2 + (𝐡 sin πœƒ)2

𝑅2 = 𝐴2 + (𝐡 cos πœƒ)2 + 2𝐴𝐡 cos πœƒ + 𝐡2 sin2πœƒ

𝑅2 = 𝐴2 + 𝐡2 cos2πœƒ + 2𝐴𝐡 cos πœƒ + 𝐡2 sin2 πœƒ

𝑅2 = 𝐴2 + 𝐡2 (cos2πœƒ + sin2 πœƒ) + 2𝐴𝐡cosπœƒ

𝑅2 = 𝐴2 + 𝐡2 + 2𝐴𝐡cos πœƒ

𝑹 = √(π‘¨πŸ + π‘©πŸ + πŸπ‘¨π‘© 𝐜𝐨𝐬𝜽)

Direction: Let 𝛼 be angle made by the resultant vector 𝑅⃗ with the vector 𝐴⃗, then tan 𝛼 = 𝑆𝑁/𝑂𝑁 = 𝑆𝑁/(𝑂𝑃 + 𝑃𝑁)

tan 𝛼 = 𝐡sinπœƒ/(𝐴 + 𝐡cosπœƒ)

𝜢 = π­πšπ§βˆ’πŸ(π‘©π¬π’π§πœ½/(𝑨 + π‘©πœπ¨π¬πœ½)

Limitations of Graphical method of adding vectors

(i) It is very difficult method.

(ii) It has limited accuracy.

To overcome these limitations Analytical method of addition of vectors is preferred.

Addition of vectors – Analytical method

In two Dimensions

Consider two vectors π‘Žβƒ— and 𝑏⃗ in X-Y plane. If π‘Žβƒ— = π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘— Μ‚ and 𝑏⃗ = 𝑏π‘₯𝑖̂+ 𝑏𝑦𝑗 Μ‚ then, 𝑅⃗ = π‘Žβƒ— + 𝑏⃗

𝑅⃗ = (π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚) + (𝑏π‘₯𝑖̂+ 𝑏𝑦𝑗̂)

𝑅⃗ = π‘Žπ‘₯𝑖̂+ 𝑏π‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚+ 𝑏𝑦𝑗 Μ‚

𝑅⃗ = (π‘Žπ‘₯ + 𝑏π‘₯ )𝑖̂+ (π‘Žπ‘¦ + 𝑏𝑦)𝑗̂

𝑹⃗ = π‘Ήπ’™π’ŠΜ‚+ π‘Ήπ’šπ’‹ Μ‚

Where 𝑅π‘₯ = π‘Žπ‘₯ + 𝑏π‘₯ and 𝑅𝑦 = π‘Žπ‘¦ + 𝑏𝑦

In three Dimensions

If π‘Žβƒ— = π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚+ π‘Žπ‘§π‘˜Μ‚ and 𝑏⃗ = 𝑏π‘₯𝑖̂+ 𝑏𝑦𝑗̂+ π‘π‘§π‘˜Μ‚

then, 𝑅⃗ = π‘Žβƒ— + 𝑏⃗

𝑅⃗ = (π‘Žπ‘₯𝑖̂+ π‘Žπ‘¦π‘—Μ‚+ π‘Žπ‘§π‘˜Μ‚) + (𝑏π‘₯𝑖̂+ 𝑏𝑦𝑗̂+ π‘π‘§π‘˜Μ‚)

𝑅⃗ = π‘Žπ‘₯𝑖̂ + 𝑏π‘₯𝑖̂ + π‘Žπ‘¦π‘—Μ‚ + 𝑏𝑦𝑗̂ + π‘Žπ‘§π‘˜Μ‚ + π‘π‘§π‘˜Μ‚

𝑅⃗ = (π‘Žπ‘₯ + 𝑏π‘₯) 𝑖̂+ (π‘Žπ‘¦ + 𝑏𝑦) 𝑗̂+ (π‘Žπ‘§ + 𝑏𝑧) π‘˜Μ‚

𝑹⃗ = π‘Ήπ’™π’ŠΜ‚+ π‘Ήπ’šπ’‹Μ‚+ π‘Ήπ’›π’ŒΜ‚

This method can be extended to addition and subtraction of any number of vectors.

Motion in a plane

Position vector

The position vector π‘Ÿβƒ— of a particle located in X-Y plane with reference to the origin is given by, π‘Ÿβƒ— = π‘₯𝑖̂+ 𝑦𝑗̂

Where π‘₯ and 𝑦 are component of π‘Ÿβƒ— along X-axis and Y-axis respectively.


Consider a particle moves along curve. Initially it is at 𝑃1 at time 𝑑1 and moves to a new position 𝑃2 at time 𝑑2. Then the displacement is given by, βˆ†π‘Ÿβƒ— = π‘Ÿβƒ—2 βˆ’ π‘Ÿβƒ—1

βˆ†π‘Ÿβƒ— = (π‘₯2𝑖̂+ 𝑦2𝑗̂) βˆ’ (π‘₯1𝑖̂+ 𝑦1𝑗̂)

βˆ†π‘Ÿβƒ— = (π‘₯2π‘–Μ‚βˆ’π‘₯1𝑖)Μ‚ + (𝑦2π‘—Μ‚βˆ’ 𝑦1𝑗̂)

βˆ†π‘Ÿβƒ— = (π‘₯2βˆ’π‘₯1) 𝑖̂ + (𝑦2 βˆ’ 𝑦1) 𝑗̂

βˆ†π’“βƒ— = βˆ†π’™π’ŠΜ‚+ βˆ†π’šπ’‹Μ‚

Where βˆ†π‘₯ = π‘₯2 βˆ’ π‘₯1 and βˆ†π‘¦ = 𝑦2 βˆ’ 𝑦1


Average velocity

It is defined as ratio of the displacement to the time taken.

𝑣⃗̅= βˆ†π‘Ÿβƒ—/βˆ†π‘‘ = (βˆ†π‘₯𝑖̂+ βˆ†π‘¦π‘—Μ‚)/βˆ†π‘‘

𝑣⃗̅= (βˆ†π‘₯/βˆ†π‘‘) 𝑖̂+ (βˆ†π‘¦/βˆ†π‘‘) 𝑗̂

𝒗⃗̅ = π’—Μ…π’™π’ŠΜ‚+ π’—Μ…π’šπ’‹Μ‚

Where 𝒗̅𝒙 = βˆ†π‘₯/βˆ†π‘‘ and π’—Μ…π’š = βˆ†π‘¦/βˆ†π‘‘

Direction of the average velocity is same as that of the displacement.

Instantaneous velocity (Velocity)

It is given by the limiting value of the average velocity as the time interval approaches to zero.

𝑣⃗ = lim βˆ†π‘‘β†’0 βˆ†π‘Ÿβƒ—/βˆ†π‘‘

𝒗⃗ = 𝒅𝒓⃗/𝒅𝒕

The direction of velocity at any point on the path of the object is tangential to the path at that point and in the direction of the motion. The components of the velocity 𝑣⃗ are given by, 𝑣⃗ = lim βˆ†π‘‘β†’0 βˆ†π‘Ÿβƒ—/βˆ†π‘‘

𝑣⃗ = lim βˆ†π‘‘β†’0 (βˆ†π‘₯/βˆ†π‘‘) 𝑖̂+ (βˆ†π‘¦/βˆ†π‘‘) 𝑗̂

𝑣⃗ = lim βˆ†π‘‘β†’0 (βˆ†π‘₯/βˆ†π‘‘) 𝑖̂ + lim βˆ†π‘‘β†’0 (βˆ†π‘¦/βˆ†π‘‘) 𝑗̂

𝑣⃗ = (𝑑π‘₯/𝑑𝑑) 𝑖̂+ (𝑑𝑦/𝑑𝑑) 𝑗̂

𝒗⃗ = π’—π’™π’ŠΜ‚+ π’—π’šπ’‹Μ‚

The magnitude is given by 𝒗 = √(π’—π’™πŸ + π’—π’šπŸ) and Direction is given by

𝜽 = π­πšπ§βˆ’πŸ(π’—π’š/𝒗𝒙)


Average acceleration

It is defined as the change in velocity divided by time interval.

π‘ŽΜ…βƒ— = βˆ†π‘£βƒ—/βˆ†π‘‘ = βˆ†(𝑣π‘₯𝑖̂+𝑣𝑦𝑗̂)/βˆ†π‘‘

π‘ŽΜ…βƒ— = (βˆ†π‘£π‘₯/βˆ†π‘‘) 𝑖̂ + (βˆ†π‘£π‘¦/βˆ†π‘‘) 𝑗̂

𝒂⃗̅⃗ = π’‚Μ…π’™π’ŠΜ‚+ π’‚Μ…π’šπ’‹Μ‚

Instantaneous acceleration (Acceleration)

It is the limiting value of the average acceleration as the time interval approaches zero.

π‘Žβƒ— = limβˆ†π‘‘β†’0 (βˆ†π‘£βƒ—/βˆ†π‘‘)

𝒂⃗ = 𝒅𝒗⃗/𝒅𝒕

The Components are given by, π‘Žβƒ— = lim βˆ†π‘‘β†’0 (βˆ†π‘£βƒ—/βˆ†π‘‘)

π‘Žβƒ— = lim βˆ†π‘‘β†’0 βˆ†(𝑣π‘₯𝑖̂+ 𝑣𝑦𝑗̂)/βˆ†π‘‘

π‘Žβƒ— = lim βˆ†π‘‘β†’0 (βˆ†π‘£π‘₯/βˆ†π‘‘) 𝑖̂+ lim βˆ†π‘‘β†’0 (βˆ†π‘£π‘¦/βˆ†π‘‘) 𝑗̂

π‘Žβƒ— = (𝑑𝑣π‘₯/𝑑𝑑) 𝑖̂+ (𝑑𝑣𝑦/𝑑𝑑) 𝑗̂

𝒂⃗ = π’‚π’™π’ŠΜ‚+ π’‚π’šπ’‹Μ‚

In one dimension the direction of velocity and acceleration is same or in opposite direction but in two or three dimensions, velocity and acceleration vectors may have any angle between 0Β° and 180Β°.

Motion in a plane with constant acceleration

Consider an object moving in X-Y plane and its acceleration π‘Žβƒ— is constant. Let the velocity of the object be 𝑣⃗0 at time 𝑑 = 0 and 𝑣⃗ at time 𝑑, then

(i) 𝑣⃗ = 𝑣⃗0 + π‘Žβƒ—π‘‘

In terms of its components,

𝑣π‘₯ = 𝑣0π‘₯ + π‘Žπ‘₯𝑑

𝑣𝑦 = 𝑣0𝑦 + π‘Žπ‘¦π‘‘

(ii) Displacement is π‘Ÿβƒ— βˆ’ π‘Ÿβƒ—0 = 𝑣⃗0𝑑 + (1/2)π‘Žβƒ—π‘‘2

In terms of its components,

π‘₯ βˆ’ π‘₯0 = 𝑣0π‘₯𝑑 + (1/2)π‘Žπ‘₯𝑑2

𝑦 βˆ’ 𝑦0 = 𝑣0𝑦𝑑 + (1/2)π‘Žπ‘¦π‘‘2

(iii) 𝑣 βƒ—2 = 𝑣0 βƒ—2 + 2π‘Žβƒ—(π‘Ÿβƒ— βˆ’ π‘Ÿβƒ—0)

In terms of its components,

𝑣π‘₯2 = 𝑣0π‘₯2 + 2π‘Žπ‘₯(π‘₯ βˆ’ π‘₯0)

𝑣𝑦2 = 𝑣0𝑦2 + 2π‘Žπ‘¦(𝑦 βˆ’ 𝑦0)

The motion in plane can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.

Relative velocity in two Dimensions

Suppose two objects 𝐴 and 𝐡 are moving with velocities 𝑣⃗𝐴 and 𝑣⃗𝐡, then the velocity of the object 𝐴 relative to that of 𝐡 is, 𝑣⃗𝐴𝐡 = 𝑣⃗𝐴 βˆ’ 𝑣⃗𝐡.

Similarly, the velocity of the object 𝐡 relative to that of 𝐴 is, 𝑣⃗𝐡𝐴 = 𝑣⃗𝐡 βˆ’ 𝑣⃗𝐴 Therefore, 𝑣⃗𝐴𝐡 = βˆ’π‘£βƒ—π΅π΄ and |𝑣⃗𝐴𝐡| = |𝑣⃗𝐡𝐴|

Examples for Motion in a plane

(i) Projectile motion (Uniformly accelerated motion)

(ii) Circular motion (non-uniformly accelerated motion)

When a particle traces a curve in two-dimensional plane, the velocity of the particle changes at least in direction. Hence, a two-dimensional motion along a curve is essentially an accelerated motion. Acceleration may be uniform or non-uniform.


A projectile is any object thrown into air or space.

Projectile motion

Motion associated with a projectile in parabolic path is called Projectile motion. Ex: A ball leaving the hand of a bowler, A stone thrown at an angle to the horizontal, an object dropped from an aeroplane in horizontal flight. The motion of projectile may be thought of as the result of two separate, simultaneously occurring components of motion. One component is along a horizontal direction with any acceleration and other along the vertical direction with constant acceleration due to gravity. It was Galileo, who first stated this independency of the horizontal and vertical components of projectile motion.

Analysis of Projectile motion

Let a projectile is projected with initial velocity 𝑣⃗0 that makes an angle πœƒ with x-axis. The acceleration acting on it is due to gravity and is directed vertically downwards. π‘Žπ‘₯ = 0, π‘Žπ‘¦ = βˆ’π‘”, hence π‘Žβƒ— = βˆ’π‘”π‘—Μ‚.

The components of initial velocity 𝑣⃗0 are, 𝑣0π‘₯ = 𝑣0cosπœƒ, 𝑣0𝑦 = 𝑣0sinπœƒ

The components of velocity at time 𝑑 are, 𝑣π‘₯ = 𝑣0 cosπœƒ, 𝑣𝑦 = 𝑣0 sin πœƒ – 𝑔𝑑.

The components of displacements at time 𝑑 are,

(i) π‘₯ = 𝑣0π‘₯𝑑 + (1/2)π‘Žπ‘₯𝑑2

π‘₯ = 𝑣0π‘₯𝑑 (∡ π‘Žπ‘₯ = 0), 𝒙 = (π’—πŸŽ 𝐜𝐨𝐬 𝜽) 𝒕 (along X-axis)

(ii) 𝑦 = 𝑣0𝑦𝑑 + (1/2)π‘Žπ‘¦π‘‘2, π’š = (π’—πŸŽ 𝐬𝐒𝐧𝜽)𝒕 – (𝟏/𝟐)π’ˆπ’•πŸ (along π‘Œ βˆ’ axis)

Path of a projectile

The path described by the projectile is called trajectory. The trajectory is a parabola.

Expression for path of a projectile (Show that the path of a projectile is Parabola)

The displacement of the projectile along X-axis is, π‘₯ = (𝑣0 cos πœƒ)𝑑, 𝑑 = π‘₯/𝑣0cos πœƒ The displacement of the projectile along Y-axis is,

𝑦 = (𝑣0 sin πœƒ)𝑑 – (1/2)𝑔𝑑2

𝑦 = (𝑣0 sin πœƒ) (π‘₯/𝑣0 cos πœƒ) – (1/2)𝑔(π‘₯/𝑣0cos πœƒ)2

𝑦 = (sin πœƒ/cos πœƒ)π‘₯ – (1/2)(𝑔/𝑣02 cos2πœƒ)π‘₯2

𝑦 = π‘₯ tan πœƒ – (1/2)(𝑔/𝑣02cos2πœƒ)π‘₯2

π’š = 𝒂𝒙 βˆ’ π’ƒπ’™πŸ

where π‘Ž = π‘‘π‘Žπ‘›πœƒ and 𝑏 = (1/2)𝑔𝑣02π‘π‘œπ‘ 2πœƒ

The equation 𝑦 = π‘Žπ‘₯ βˆ’ 𝑏π‘₯2 represents a parabola.

Hence the trajectory is a parabola.

Time of Flight

It is the time during which the projectile is in flight. It is denoted by 𝑇𝑓.

Expression for Time of flight

The component of velocity along Y-axis at time t is, 𝑣𝑦 = 𝑣0 sinπœƒ βˆ’ 𝑔𝑑

At maximum height 𝑣𝑦 = 0 and time for maximum height, 𝑑 = π‘‘π‘š.

0 = 𝑣0 sin πœƒ βˆ’ π‘”π‘‘π‘š

π‘”π‘‘π‘š = 𝑣0 sinπœƒ

π‘‘π‘š = 𝑣0 sinπœƒ/𝑔

Time of flight 𝑇𝑓 = 2π‘‘π‘š because β€œtime of ascent = time of descent”

𝑻𝒇 = πŸπ’—πŸŽπ¬π’π§πœ½/π’ˆ

Maximum height of a projectile

It is the maximum height reached by the projectile in time π‘‘π‘š. It is denoted by β„Žπ‘š.

Expression for maximum height of a projectile

The displacement along Y βˆ’ axis is, 𝑦 = (𝑣0 sinπœƒ)π‘‘π‘š – (1/2)π‘”π‘‘π‘š2

β„Žπ‘š = (𝑣0 sin πœƒ)(𝑣0 sin πœƒ/𝑔) – (1/2)𝑔(𝑣0sin πœƒ/𝑔)2

β„Žπ‘š = (𝑣0 sin πœƒ)2/𝑔 – (1/2)𝑔((𝑣0 sinπœƒ)2/𝑔2)

β„Žπ‘š = (𝑣0 sin πœƒ)2/𝑔 – (1/2)(𝑣0 sinπœƒ)2𝑔

β„Žπ‘š = (1 – (1/2))(𝑣0 sinπœƒ)2/𝑔

β„Žπ‘š = (1/2)(𝑣0 sinπœƒ)2/𝑔

π’‰π’Ž = (π’—πŸŽπ¬π’π§πœ½)𝟐/πŸπ’ˆ

Horizontal Range of projectile

It is the horizontal distance covered by the projectile during its flight. It is denoted by 𝑅.

Expression for Horizontal Range of projectile

Displacement along X-axis is, π‘₯ = (𝑣0 cosπœƒ)𝑑

Now π‘₯ = 𝑅 and 𝑑 = 𝑇𝑓

𝑅 = (𝑣0 cosπœƒ)𝑇𝑓

𝑅 = (𝑣0 cosπœƒ)(2𝑣0 sinπœƒ/𝑔)

𝑅 = (𝑣02/𝑔)(2cosπœƒsinπœƒ)

𝑹 = (π’—πŸŽπŸ/π’ˆ)𝐬𝐒𝐧𝟐𝜽


(i) For a given speed of projection, the projectile will have a maximum range (π‘…π‘š) when sin 2πœƒ is maximum or the angle of projection is 45Β°. sin 2πœƒ = 1 ⟹ 2πœƒ = 90Β° Then angle of projection, πœƒ = 45Β°.

Maximum range, π‘…π‘š = (𝑣02/𝑔) sin2(45Β°)

π‘Ήπ’Ž = π’—πŸŽπŸ/π’ˆ

(ii) Show that π‘Ήπ’Ž = πŸ’π’‰π’Ž when the angle of projection is 𝜽 = πŸ’πŸ“Β°.

For πœƒ = 45Β°, π‘…π‘š = 𝑣02/𝑔 and β„Žπ‘š = 𝑣02sin2πœƒ/2𝑔 = 𝑣02sin245Β°/2𝑔 = (𝑣02/2𝑔) Γ— (1/2) β„Žπ‘š = 𝑣02/4𝑔 = (1/4)(𝑣02/𝑔)

β„Žπ‘š = (1/4)(π‘…π‘š)

π‘Ήπ’Ž = πŸ’π’‰π’Ž

Since the maximum range of a projectile is equal to 4 times the maximum height reached.

Uniform circular motion

The motion of the object in a circular path at a constant speed is called uniform circular motion. Even though the object moves at a constant speed it has acceleration, because there is a continuous change in its direction of motion. Hence there is a change in its velocity from point to point.

Expression for Acceleration

Let π‘Ÿβƒ— and π‘Ÿβƒ—β€² be the position vectors and 𝑣⃗ and 𝑣⃗′ are the velocities of the object when it is at 𝑃 and 𝑄 as shown. Velocity at a point is along the tangent at that point in the direction of motion. From βˆ†π΄π΅πΆ, 𝐴𝐡⃗ + 𝐡𝐢⃗ = 𝐴𝐢⃗

𝑣⃗ + 𝐡𝐢⃗ = 𝑣⃗′

𝐡𝐢⃗ = 𝑣⃗′ βˆ’ 𝑣⃗ = βˆ†π‘£βƒ—

βˆ†π‘£βƒ— is the change in velocity, which is towards the center. Since the path is circular, 𝑣⃗ and 𝑣⃗′ are perpendicular to π‘Ÿβƒ— and π‘Ÿβƒ—β€² respectively. Therefore βˆ†π‘£βƒ— is perpendicular to βˆ†π‘Ÿβƒ—. The average acceleration = βˆ†π‘£βƒ—/βˆ†π‘‘.

Since βˆ†π‘£βƒ— is perpendicular to βˆ†π‘Ÿβƒ—, π‘ŽΜ… is along βˆ†π‘£βƒ— and perpendicular to βˆ†π‘Ÿβƒ— and directed towards the center of the circle. The Instantaneous acceleration is,

π‘Žβƒ— = lim βˆ†π‘‘β†’0 βˆ†π‘£βƒ—/βˆ†π‘‘

Its magnitude is given by,

π‘Žπ‘ = |π‘Žβƒ—|

π‘Žπ‘ = lim βˆ†π‘‘β†’0 |βˆ†π‘£βƒ—|/βˆ†π‘‘

Since the velocity vectors 𝑣⃗ and 𝑣⃗′ are always perpendicular to π‘Ÿβƒ— and π‘Ÿβƒ—β€² , the angle between 𝑣⃗ and 𝑣⃗′ is also βˆ†πœƒ. Since βˆ†π΄π΅πΆ and βˆ†π‘‚π‘ƒπ‘„ are similar.

Then, 𝐡𝐢/𝑃𝑄 = 𝐴𝐡/𝑂𝑃

|βˆ†π‘£βƒ—|/|βˆ†π‘Ÿβƒ—| = 𝑣/𝑅 (∡ |π‘Ÿβƒ—| = 𝑅 = π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ )

βˆ†π‘£βƒ— = (|βˆ†π‘Ÿβƒ—|𝑣)/𝑅

∴ π‘Žπ‘ = lim βˆ†π‘‘β†’0 (|βˆ†π‘Ÿβƒ—|𝑣/βˆ†π‘‘π‘…)

π‘Žπ‘ = (𝑣/𝑅)lim βˆ†π‘‘β†’0 |βˆ†π‘Ÿβƒ—|/βˆ†π‘‘

π‘Žπ‘ = (𝑣/𝑅)𝑣

𝒂𝒄 = π’—πŸ/𝑹

This equation represents the magnitude of acceleration and is directed toward the center.

Centripetal acceleration

The acceleration, which is directed towards the center, is called centripetal acceleration. The term centripetal acceleration was termed by Newton and Centripetal comes from a Greek term that means Centre seeking towards the center.


(i) In uniform circular motion as the object moves from P to Q (in the above figure) in time βˆ†π‘‘ the line OP turns through an angle βˆ†πœƒ, called angular distance. But angular speed, πœ” = βˆ†πœƒ/βˆ†π‘‘ If the distance traveled 𝑃𝑄 = βˆ†π‘  then, Speed, 𝑣 = βˆ†π‘ /βˆ†π‘‘ But βˆ†π‘  = π‘…βˆ†πœƒ, where R is the radius of the trajectory.

∴ 𝑣 = π‘…βˆ†πœƒ/βˆ†π‘‘ = 𝑅(βˆ†πœƒ/βˆ†π‘‘)

𝒗 = π‘ΉπŽ β†’ (1)

The Centripetal acceleration π‘Žπ‘ = 𝑣2/𝑅 = 𝑅2πœ”2/𝑅

𝒂𝒄 = π‘ΉπŽπŸ

(ii) Time period (T)

Time is taken by an object to make one revolution.

(iii)Frequency (𝝂)

The number of revolutions made in one second. 𝜈 = 1/𝑇.

Distance moved in time period 𝑇 = 2πœ‹π‘…

Speed, 𝑣 = 2πœ‹π‘…/𝑇 = 2πœ‹π‘…πœˆ

𝑣 = 𝑅(2πœ‹πœˆ) β†’ (2)

Comparing equations (1) and (2), we get 𝝎 = πŸπ…π‚

Then Acceleration, π‘Žπ‘ = (2πœ‹πœˆ)2𝑅

𝒂𝒄 = πŸ’π…πŸπ‚πŸπ‘Ή


1) A cricket ball is thrown at a speed of 28π‘šπ‘ βˆ’1 in a direction of 30Β° with the horizontal.


(a) the maximum height

(b) the time taken by the ball to return to the same level.

(c) the distance from the thrower to the point where the ball returns to the same level.

Given 𝑣0 = 28π‘šπ‘ βˆ’1, πœƒ = 30Β°

(a) β„Žπ‘š = (𝑣0 sin πœƒ)2/2𝑔

β„Žπ‘š = (28 Γ— sin 30Β°)2/(2 Γ— 9.8)

β„Žπ‘š = (28 Γ— (1/2)2/(2 Γ— 9.8)

β„Žπ‘š = 142/19.6

β„Žπ‘š = 10π‘š

(b) 𝑇𝑓 = 2𝑣0 sin πœƒ/𝑔

𝑇𝑓 = (2 Γ— 28 Γ— sin 30Β°)/9.8 = (2 Γ— 28 Γ— (1/2))/9.8

𝑇𝑓 = 2.85𝑠

(c) 𝑅 = (𝑣02/𝑔)sin2πœƒ

𝑅 = (282/9.8)sin(2 Γ— 30Β°) = (282/9.8) Γ— 0.866

𝑅 = 69.28π‘š

2) A cricket ball projected at an angle of 30Β° with the horizontal takes 3 seconds to reach the ground.


(a) the velocity of the projection.

(b) The horizontal range of the ball.

Given, = 30Β°, 𝑇𝑓 = 3𝑠

(π‘Ž) 𝑇𝑓 = 2𝑣0 sin πœƒ/𝑔

3 = (2 Γ— 𝑣0 Γ— sin 30Β°/9.8)

𝑣0 = (3 Γ— 9.8)/(2 Γ— sin 30Β°) = (3 Γ— 9.8)/(2 Γ— (1/2))

𝑣0 = 29.4π‘šπ‘ βˆ’1

(𝑏) 𝑅 = (𝑣02/𝑔)sin2πœƒ

𝑅 = ((29.4)2/9.8) sin(2 Γ— 30Β°) = (864.36/9.8) Γ— 0.866

𝑅 = 76.38π‘š

3) The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40π‘šπ‘ βˆ’1 can go without hitting the ceiling of the hall?

Given 𝑣0 = 40π‘šπ‘ βˆ’1 , Height of the hall, β„Žπ‘š = 25π‘š, πœƒ = ?, β„Žπ‘š = (𝑣0 sin πœƒ)2/2𝑔

25 = (40 Γ— sin πœƒ)2/(2 Γ— 9.8)

25 Γ— 2 Γ— 9.8 = (40 Γ— sin πœƒ)2

40 Γ— sin πœƒ = √(25 Γ— 2 Γ— 9.8)

sin πœƒ = √(25 Γ— 2 Γ— 9.8)/40 = 0.5534

πœƒ = sinβˆ’1(0.5534) = 33.6Β°

𝑅 = (𝑣02/𝑔)sin 2πœƒ

𝑅 = (402/9.8) sin(2 Γ— 33.6Β°)

𝑅 = (1600/9.8) Γ— 0.9219

𝑅 = 150.5π‘š

4) A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball?

Given π‘…π‘š = 100π‘š,

For maximum range, πœƒ = 45Β°

π‘…π‘š = 𝑣02/𝑔

100 = 𝑣02/9.8

𝑣02 = 100 Γ— 9.8 = 980

𝑣0 = 31.3π‘šπ‘ βˆ’1

To throw the ball vertically upwards, πœƒ = 90Β°

β„Žπ‘š = (𝑣0 sin πœƒ)2/2𝑔

β„Žπ‘š = (31.3 Γ— sin 90Β°)2/(2 Γ— 9.8)

β„Žπ‘š = (31.3 Γ— 31.3)(2 Γ— 9.8)

β„Žπ‘š = 49.98π‘š β‰ˆ 50π‘š

5) A stone tide to the end of a string 80cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25s, what is the magnitude and direction of acceleration of the stone?

Given Radius, π‘Ÿ = 80π‘π‘š = 0.80π‘š

Frequency, 𝜈 = 14/25 π‘ βˆ’1

πœ” = 2πœ‹πœˆ

πœ” = 2 Γ— (22/7) Γ— (14/25)

πœ” = 3.52 π‘Ÿπ‘Žπ‘‘/𝑠

π‘Žπ‘ = π‘Ÿπœ”2

π‘Žπ‘ = 0.80 Γ— (3.52)2

π‘Žπ‘ = 9.91π‘šπ‘ βˆ’2

6) An aircraft executes a horizontal loop of radius 1km with a speed of 900kmph. Compare its centripetal acceleration with the acceleration due to gravity.

Given Radius, π‘Ÿ = 1π‘˜π‘š = 1000π‘š

Speed, 𝑣 = 900π‘˜π‘šπ‘β„Ž = 900 Γ— (1000/3600)π‘šπ‘ βˆ’1 = 250π‘šπ‘ βˆ’1

π‘Žπ‘ = 𝑣2/π‘Ÿ

π‘Žπ‘ = 2502/1000 = 62500/1000

π‘Žπ‘ = 62.5π‘šπ‘ βˆ’2

Comparison: But 𝑔 = 9.8π‘šπ‘ βˆ’2

π‘Žπ‘/𝑔 = 62.5/9.8 = 6.38

π‘Žπ‘ = 6.38𝑔

7)Rain is falling vertically with a speed of 30π‘šπ‘ βˆ’1. A woman rides a bicycle with a speed of 10π‘šπ‘ βˆ’1 in the north-south direction. What is the direction in which she should hold her umbrella?

Given Velocity of bicycle, 𝑣⃗𝐡 = 10π‘šπ‘ βˆ’1

Velocity of rain, 𝑣⃗𝑅 = 30π‘šπ‘ βˆ’1

Magnitude of velocity, 𝑣𝑅𝐡 = √(𝑣𝐡2 + 𝑣𝑅2)

𝑣𝑅𝐡 = √(302 + 102) = √1000

𝑣𝑅𝐡 = 31.6π‘šπ‘ βˆ’1

Direction, πœƒ = tanβˆ’1(𝑣𝐡/𝑣𝑅)

πœƒ = tanβˆ’1 (10/30) = tanβˆ’1(0.3333)

πœƒ = 18Β°26β€² She should hold the umbrella at an angle 18Β°26β€² with the vertical in south-west direction.

8) An aeroplane flying at 540kmph drops a missile towards the ground. If the height of the plane is 1000m then calculate

(a) time taken by the missile to hit the ground.

(b) Horizontal distance covered by the missile from the initial point.

Given π‘Žπ‘¦ = 9.8π‘šπ‘ βˆ’2, π‘Žπ‘₯ = 0, 𝑦 = 1000π‘š.

At the top, 𝑣0π‘₯ = 540π‘˜π‘šπ‘β„Ž = 150π‘šπ‘ βˆ’1, 𝑣0𝑦 = 0

(a) Time taken can be calculated as, 𝑦 = 𝑣0𝑦𝑑 + (1/2)π‘Žπ‘¦π‘‘2

1000 = 0 + (1/2) Γ— 9.8 Γ— 𝑑2

𝑑2 = (1000 Γ— 2)/9.8

𝑑 = 14.29𝑠

(b) π‘₯ = 𝑣0π‘₯𝑑 + (1/2)π‘Žπ‘₯𝑑2

π‘₯ = 150 Γ— 14.29

π‘₯ = 2142.86π‘š

9) Two concurrent forces 20N and 30N are acting at an angle of 60Β° with respect to each other. Calculate the magnitude and direction of the resultant. Given 𝐴 = 20𝑁, 𝐡 = 30𝑁, πœƒ = 60Β°

(a) Magnitude, 𝑅 = √(𝐴2 + 𝐡2 + 2𝐴𝐡 cosπœƒ)

𝑅 = √(202 + 302 + (2 Γ— 20 Γ— 30 Γ— cos 60Β°)

𝑅 = √(400 + 900 + (2 Γ— 600 Γ— (1/2)) = √(1900)

𝑅 = 43.50𝑁

(b) Direction, 𝛼 = tanβˆ’1(𝐡 sinπœƒ/(𝐴 + 𝐡 cos πœƒ))

𝛼 = tanβˆ’1(30 Γ— sin 60Β°/(20 + (30 Γ— cos 60Β°))

𝛼 = tanβˆ’1(25.98/35) = 36Β°58β€²

The resultant force is 43.50𝑁 in the direction of 36Β°58β€² with respect to 20𝑁.

Important Questions.

One mark.

1) What is unit vector?

2) What is Zero (null) vector?

3) Is scalar multiplied by a vector, a vector or a scalar?

4) What is the minimum number of vectors to give zero resultant?

5) When will be the resultant of two given vectors is maximum?

6) What is resolution of vector?

7) What is time of flight of a projectile?

8) At what angle range of a projectile is maximum? or when the range of a projectile does become maximum?

9) What is the relation between maximum height and maximum range of a projectile?

10) For angle of projection 30Β°, 𝑅 is the range of the projectile. Then write another angle of projection for which the range is same.

11) Represent the unit vector in mathematical form.

Two marks.

1) What are scalar and vector? Give example. OR distinguish between scalar and vector.

2) What is a projectile? Give an example.

3) Write the equation for the trajectory of a projectile motion. What is the nature of its trajectory?

4) State and explain parallelogram law of vector addition.

5) A unit vector is represented by 𝐴𝑖̂+ 𝐡𝑗̂+ πΆπ‘˜Μ‚. If the value of 𝐴 and 𝐡 are 0.5 and 0.8 respectively, then find the value of 𝐢.

Three marks.

1) State and explain the triangle law of vectors addition.

2) Obtain an expression for maximum height reached by a projectile.

3) Obtain an expression for time of flight of a projectile.

4) What is resolution of vectors? Write expressions for π‘₯ and 𝑦 components (Rectangular) of a vector. or obtain the equations for rectangular components of a vector in two dimensions.

5) Derive an expression for magnitude of resultant of two concurrent vectors. or find the magnitude of the resultant of two vectors 𝐴 and 𝐡 in terms of their magnitude and angle πœƒ between them.

6) Obtain the expression for range of a projectile.

Five marks

1) What is centripetal acceleration? Derive an expression for centripetal acceleration of a particle in uniform circular motion. or What is centripetal acceleration? Derive the expression for radial acceleration.

2) What is projectile motion? Show that trajectory of projectile is a parabola. or What is projectile motion? Derive an expression for trajectory of projectile. or show that the path of the projectile is a parabola.

Additional Problems

1) A bullet is fired at a velocity of 392π‘šπ‘ βˆ’1 at an angle of 30Β° to the horizontal. Find the maximum height attended ad time of flight.

2) A body is projected with a velocity of 50π‘šπ‘ βˆ’1 in a direction making an angle of 30Β° with the horizontal.


(a) The maximum height.

(b) the time taken by the body to return to the same level

and (c) the range.

3) A ball is thrown into air with a speed of 62π‘šπ‘ βˆ’1 at an angle 45Β° with the horizontal.


(a) The maximum height attained.

(b) the time of flight.

(c) the horizontal range.

4) A player hits a cricket ball at angle of 40Β° to the horizontal. If the ball moves with a velocity of 20π‘šπ‘ βˆ’1. Find

(a) The maximum height reached by the ball.

(b) the time of flight.

(c) the horizontal range.

Given 𝑔 = 10π‘šπ‘ βˆ’1.

5) A football player kicks a ball at an angle of 30Β° to the horizontal with an initial velocity of 15π‘šπ‘ βˆ’1. Assuming that the ball travels in a vertical plane. Calculate (a) The maximum height reached. (b) the time of flight and (c) the horizontal range.

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