CLASS XI PHYSICS TOPIC NO.4

1. Motion in a Plane

2. • Distance = 20cm
• Speed =20m/s
• Time= 60s
• Temperature = 200K
• Displacement = 20cm east
• velocity =40 m/s north
• Acceleration = 30m/s south
• Force = 20N east
• Numerical value (magnitude)
• Unit
• Numerical value (magnitude)
• Unit
• Direction
Scalar Quantities Vector Quantities

3. Scalars and Vectors
Scalars Vectors
1. Scalars have only magnitude. 1. Vectors have both magnitude and
direction.
1. They change if their magnitude
changes.
2. They change if either their magnitude,
direction or both change.
2. They can be added according to
ordinary laws of algebra.
ex: Distance, speed, work, mass,
density, etc
3. They can be added only by using laws
of vector addition.
ex: Displacement, velocity, force, etc

4. Vectors:
Those physical quantities which require magnitude as well as direction for
their complete representation and follows vector laws are called vectors
or vector quantities. A vector quantity is specified by a number with a unit
and its direction.
Representation of a vector:
A vector quantity is represented by a straight line arrowhead over it.
ex: i) force vector is represented as 𝐹 or F.

5. Position and Displacement vectors:
Position vector: a vector which gives position of an object with reference
to the origin of a co-ordinate system.
Displacement vector:
it is that vector which tells how much and in which direction an object has
changed its position in a given time interval.

6. Different Types of Vectors:
Equal Vectors:
Two vectors are said to be equal if
they have same magnitude and
same direction.
𝐴 and 𝐵 are
equal vectors.
Negative of a Vector:
The negative of a vector is defined as
another vector having the same magnitude
but having an opposite direction.
𝐵 is a negative equal
vector of 𝐴 or vice
versa
𝐴
𝐴
𝐵
𝐵

7. Modulus of a Vector:
The modulus of a vector means the length or the magnitude of the
vector.
It is the scalar quantity.
| 𝐴| = 𝐴
Unit Vector:
A unit vector is a vector of unit magnitude drawn in the direction of a given
vector.
𝑨 =
𝑨
|𝑨|
=
𝑨
𝑨

8. Collinear Vectors:
Vectors having equal or unequal magnitudes but acting along the same
or parallel lines.
Co-initial Vectors:
Vectors having a common initial point.
Co-terminus Vectors:
Vectors having a common terminal point.
Zero/ null Vector:
Vectors having zero magnitude and arbitrary direction.
represented as 0

9. Properties of Zero/ null Vector:
 when a vector is added to a zero vector, we get the same
vector.
𝐴 + 0 = 𝐴
 when a vector is multiplied to a zero vector, we get the
zero vector.
𝐴 0 = 0

10. Laws of Addition of Vectors:
1. Triangle Law of Vector Addition
If two vectors acting at a point are represented in
magnitude and direction by the two sides of a
triangle taken in one order, then their resultant is
represented by the third side of the triangle taken
in the opposite order.
If two vectors 𝐴 and 𝐵 acting at a point are
inclined at an angle θ, then their resultant 𝑅
𝑹 = 𝑨 + 𝑩
Magnitude, 𝑹 = 𝑨 𝟐 + 𝟐𝑨𝑩𝒄𝒐𝒔𝜽 + 𝑩 𝟐
𝒕𝒂𝒏𝜷 =
𝑩𝒔𝒊𝒏𝜽
(𝑨 + 𝑩𝒄𝒐𝒔𝜽)
𝑨
𝑩
𝑹

11. 2. Parallelogram Law of Vector Addition
If two vectors acting at a point are represented
in magnitude and direction by the two adjacent
sides of a parallelogram draw from a point, then
their resultant is represented in magnitude and
direction by the diagonal of the parallelogram
draw from the same point.
If two vectors 𝐴 and 𝐵 acting at a point are
inclined at an angle θ, then their resultant 𝑅
𝑹 = 𝑨 + 𝑩
Magnitude, 𝑹 = 𝑨 𝟐 + 𝟐𝑨𝑩𝒄𝒐𝒔𝜽 + 𝑩 𝟐
𝒕𝒂𝒏𝜷 =
𝑩𝒔𝒊𝒏𝜽
(𝑨+𝑩𝒄𝒐𝒔𝜽)
𝑨
𝑩
𝑹

12. 3.Polygon Law of Vector Addition
It states that if number of vectors acting on a particle
at a time are represented in magnitude and direction
by the various sides of an open polygon taken in
same order, their resultant vector E is represented
in magnitude and direction by the closing side of
polygon taken in opposite order, polygon law of
vectors is the outcome of triangle law of vectors.
𝑅 = 𝐴 + 𝐵 + 𝐶 + 𝐷 + 𝐸
𝑂𝐸= 𝑂𝐴 + 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐸

13. Properties of Vector Addition
(i) Vector addition is commutative,
i.e., 𝑨 + 𝑩 = 𝑩 + 𝑨
In fig side OP and OQ of a parallelogram OPSQ
represents vector 𝐴 and 𝐵 respectively.
Using triangle law of vector addition,
In OPS, 𝑂𝑆 = 𝑂𝑃 + 𝑃𝑆
𝑅 = 𝐴 + 𝐵
In OQS, 𝑂𝑆 = 𝑂𝑄 + 𝑄𝑆
𝑅 = 𝐵 + 𝐴
From above equations,
𝐴 + 𝐵 = 𝐵 + 𝐴
𝑨
𝑩
𝑹
O
P
S Q

14. (ii) Vector addition is associative,
i.e., 𝑨 +( 𝑩 + 𝑪 )= (𝑨 + 𝑩 )+ 𝑪
(iii) Vector addition is distributive,
i.e., m (𝑨 + 𝑩 ) = m 𝑨 + m 𝑩

15. Analytical method of vector addition:
(by using parallelogram law of vector addition)
Consider two vectors A and B making an angle θ with each other, let OP
and OQ are the sides of OPSQ represents vectors A and B respectively,
To find the magnitude of the resultant:
from point S, draw SN perpendicular to extended OP
In ONS,
OS2 =ON2 +NS2
= (OP+PN)2 +NS2
=OP2 +(2 OP* PN) +PN2 +NS2
𝑶𝑷 = 𝑨 OP= A,
𝑷𝑺 = 𝑩 PS=B,
𝑶𝑺 = 𝑹 OS=R

16. R2 = A2+2ABcosθ+B2cos2θ+B2sinθ 2
R2 = A2+2ABcosθ+B2 (cos2θ+sinθ 2)
R2 = A2+2ABcosθ+B2
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠𝜃 + 𝐵2
PN= Bcosθ ,
SN= Bsinθ

17. To find the direction of the resultant:
let the resultant R make an angle β with the
direction of A.
Then from right angled triangle ONS, we get
tan β =
SN
𝑂𝑁
=
SN
𝑂𝑃+𝑃𝑁
tan β =
Bsinθ
𝐴+𝐵𝑐𝑜𝑠θ

18. Special cases:
i) if two vectors A and B are acting along the same
direction, 𝜃=0◦
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠𝜃 + 𝐵2
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠0 + 𝐵2
R = 𝐴2 + 2𝐴𝐵 + 𝐵2
R = 𝐴 + 𝐵
tan β =
Bsinθ
𝐴+𝐵𝑐𝑜𝑠θ
=
Bsin0
𝐴+𝐵𝑐𝑜𝑠0
=0

19. ii) if two vectors A and B are acting along the opposite direction, 𝜃=180◦
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠180 + 𝐵2
R = 𝐴2 − 2𝐴𝐵 + 𝐵2
R = 𝐴 − 𝐵
tan β =
Bsinθ
𝐴+𝐵𝑐𝑜𝑠θ
=
Bsin180
𝐴+𝐵𝑐𝑜𝑠180
=0

20. i) when the two vectors A and B are acting at right angle to each other,
𝜃=90◦
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠90 + 𝐵2
R = 𝐴2 + 𝐵2
tan β =
Bsinθ
𝐴+𝐵𝑐𝑜𝑠θ
=
Bsin90
𝐴+𝐵𝑐𝑜𝑠90
=
𝐵
𝐴
𝛽 = tan−1 𝐵
𝐴

21. The flight of a bird is an example of
composition of vectors.
Working of a sling is based on law of
vector addition

22. 1. The two forces of 5N and 7N acts on a particle with an angle of 60
between them. Find the resultant force.
𝐹1= 5𝑁
𝐹2= 7𝑁
= 60
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠𝜃 + 𝐵2
F = 𝐹1
2
+ 2 𝐹1 𝐹2 𝑐𝑜𝑠𝜃 + 𝐹2
2
F = 52 + (2𝑋5𝑋7𝑐𝑜𝑠60) + 72
F = 109𝑁

23. 2. Two vectors both equal in magnitude have their resultant equal in
magnitude of the either. Find the angle between the two vectors.
A=B=R
= ?
R = 𝐴2 + 2𝐴𝐵𝑐𝑜𝑠𝜃 + 𝐵2
A = 𝐴2 + 2𝐴2 𝑐𝑜𝑠𝜃 + 𝐴2
A = 2𝐴2 + 2𝐴2 𝑐𝑜𝑠𝜃
A = 2𝐴2(1 + 𝑐𝑜𝑠𝜃)
1 + 𝑐𝑜𝑠𝜃 =
1
2
𝑐𝑜𝑠𝜃 =
1
2
− 1
𝑐𝑜𝑠𝜃 = −
1
2
𝜃=120

24. Base vectors
Base vectors are a set of vectors selected as a base to represent all
other vectors.

25. Resolution of a position vector into rectangular components:
𝑨 = 𝑨 𝒙 𝒊 + 𝑨 𝒚 𝒋 + 𝑨 𝒛 𝒌
Magnitude of vector
𝐴 = 𝐴 𝑥
2
+ 𝐴 𝑦
2
+ 𝐴 𝑧
2

26. 1. Find the magnitude of resultant of the vectors
𝑨 = 𝒊 + 𝟒 𝒋 − 2 𝒌
𝑩 = 𝟑 𝒊 − 𝟓 𝒋 + 𝒌
𝑹 = 𝑨 + 𝑩
𝑅 = ( 𝑖 + 4 𝑗 − 2 𝑘)+ (3 𝑖 − 5 𝑗 + 𝑘)
𝑅 = 4 𝑖 − 𝑗 − 𝑘
𝐴 = 𝐴 𝑥
2
+ 𝐴 𝑦
2
+ 𝐴 𝑧
2
R= 42 + (−1)2+(−1)2
R= 16 + 1 + 1
R= 18 = 3 2

27. 2. Find the magnitude of the 𝑨 + 𝑩 and 𝑨 - 𝑩
𝑨 = 𝟑 𝒊 + 𝟐 𝒋
𝑩 = 𝒊 − 𝟐 𝒋 + 𝟑 𝒌
𝑹 = 𝑨 + 𝑩
𝑅 = (3 𝑖 + 2 𝑗)+ ( 𝑖 − 2 𝑗 + 3 𝑘)
𝑅 = 4 𝑖 + 0 𝑗 + 3 𝑘
R= 42 + (0)2+(3)2
R= 16 + 9
R= 25 = 5
𝑹 = 𝑨 - 𝑩
𝑅 = (3 𝑖 + 2 𝑗) - ( 𝑖 − 2 𝑗 + 3 𝑘)
𝑅 = 3 𝑖 + 2 𝑗 - 𝑖 + 2 𝑗 − 3 𝑘)
𝑅 = 2 𝑖 + 4 𝑗 − 3 𝑘
R= (2)2+(4)2+(−3)2
R= 4 + 16 + 9
R= 29

28. Multiplication of a Vector:
By a Real Number
When a vector A is multiplied by a real number n, then its magnitude
becomes n times but direction and unit remains unchanged.
𝑩 = n 𝑨

29. Multiplication of Two Vectors
1.Scalar or Dot Product of Two Vectors
2.Vector or Cross Product of Two Vectors

30. 1. Scalar or Dot Product of Two Vectors
The scalar product of two vectors is equal to the product of their
magnitudes and the cosine of the smaller angle between them.
It is denoted by ٠ (dot).
𝑨 ٠ 𝑩 = |𝑨 ||𝑩 | cos θ = AB cos θ
cos θ =
𝑨 ٠ 𝑩
𝐴𝐵
The scalar or dot product of two vectors is a scalar.
Physical quantities such as
• W= 𝑭 ٠ 𝒔
• Power= 𝑭 ٠ 𝒗

31. Properties of Scalar Product
(i) Scalar product is commutative, i.e., A ٠ B= B ٠ A
(ii) Scalar product is distributive, i.e., A ٠ (B + C) = A ٠B + A ٠ C
(iii) Scalar product of two perpendicular vectors is zero.
A ٠ B = AB cos 90° = O
(iv) Scalar product of two parallel vectors is equal to the product of their
magnitudes,
i.e., A ٠B = AB cos 0° = AB
(v) Scalar product of a vector with itself is equal to the square of its magnitude,
i.e.,A ٠ A = AA cos 0° = A2

32. (vi) Scalar product of orthogonal unit vectors
𝒊٠ 𝒊 = 𝒋٠ 𝒋 = 𝒌 ٠ 𝒌 = 𝟏
𝒊٠ 𝒋 = 𝒋٠ 𝒌 = 𝒌 ٠ 𝒊 = 𝟎
(vii) Scalar product in cartesian coordinates
𝑨 = 𝑨 𝒙 𝒊 + 𝑨 𝒚 𝒋 + 𝑨 𝒛 𝒌
𝑩 = 𝑩 𝒙 𝒊 + 𝑩 𝒚 𝒋 + 𝑩 𝒛 𝒌
𝑨 ٠ 𝑩 = ( 𝑨 𝒙 𝒊 + 𝑨 𝒚 𝒋 + 𝑨 𝒛 𝒌)٠ ( 𝑩 𝒙 𝒊 + 𝑩 𝒚 𝒋 + 𝑩 𝒛 𝒌)
= ( 𝑨 𝒙 𝒊٠ 𝑩 𝒙 𝒊) +( 𝑨 𝒙 𝒊٠ 𝑩 𝒚 𝒋) + ( 𝑨 𝒙 𝒊٠ 𝑩 𝒛 𝒌)
+( 𝑨 𝒚 𝒋٠ 𝑩 𝒙 𝒊 )+( 𝑨 𝒚 𝒋 ٠ 𝑩 𝒚 𝒋)+( 𝑨 𝒚 𝒋 ٠ 𝑩 𝒛 𝒌)
+( 𝑨 𝒛 𝒌 ٠ 𝑩 𝒙 𝒊 )+( 𝑨 𝒛 𝒌 ٠ 𝑩 𝒚 𝒋)+( 𝑨 𝒛 𝒌 ٠ 𝑩 𝒛 𝒌)
= 𝑨 𝒙 𝑩 𝒙 + 𝑨 𝒚 𝑩 𝒚 + 𝑨 𝒛 𝑩 𝒛

33. 1. Find scalar product of
𝑨 = 𝒊 + 𝟐 𝒋 − 𝒌
𝑩 = − 𝒊 + 𝒋 − 𝟐 𝒌
𝑨 ٠ 𝑩 = ( 𝒊 + 𝟐 𝒋 − 𝒌)٠ (− 𝒊 + 𝒋 − 𝟐 𝒌)
= (1x-1) + (2x1) + (-1x-2)
= -1 + 2 + 2
= 3
𝑨 = 𝑨 𝒙 𝒊 + 𝑨 𝒚 𝒋 + 𝑨 𝒛 𝒌
𝑩 = 𝑩 𝒙 𝒊 + 𝑩 𝒚 𝒋 + 𝑩 𝒛 𝒌
𝑨 ٠ 𝑩 = 𝑨 𝒙 𝑩 𝒙 + 𝑨 𝒚 𝑩 𝒚 + 𝑨 𝒛 𝑩 𝒛

34. 2. Find the angle between the vectors,
𝑨 = 𝒊 + 𝟐 𝒋 − 𝒌 𝑎𝑛𝑑 𝑩 = − 𝒊 + 𝒋 − 𝟐 𝒌
cos θ =
𝑨 ٠ 𝑩
𝐴 𝐵
𝑨 ٠ 𝑩 = ( 𝒊 + 𝟐 𝒋 − 𝒌)٠ (− 𝒊 + 𝒋 − 𝟐 𝒌)
= (1x-1) + (2x1) + (-1x-2)
= -1 + 2 + 2
= 3
|𝑨|= 12 + 22 + (−1)2 = 1 + 4 + 1 = 6
|𝑩|= (−1)2+12 + (−2)2 = 1 + 1 + 4 = 6
cos θ =
𝟑
6 6
= =
𝟑
6
= =
𝟏
2
θ= 60
𝑨 ٠ 𝑩 = |𝑨 ||𝑩 | cos θ = AB cos θ
cos θ =
𝑨 ٠ 𝑩
𝐴 𝐵
𝐴 = 𝐴 𝑥
2
+ 𝐴 𝑦
2
+ 𝐴 𝑧
2

35. 3. A force F = (7i + 6k )N makes a body move with a velocity of
v = (3j + 4k )m/s, calculate power in watt.
F = (7i + 0j + 6k )N
v = (0i + 3j + 4k )m/s
Power = F ٠ v = (7i + 0j + 6k )٠ (0i + 3j + 4k )
= (7x0) + (0x3) + (6x4)
= 0 + 0 + 24
Power = 24watt

36. 4. Calculate the work done by a force F = (−i + 2j + 3k )N in displacing
an object through a distance of 4m along the z-axis.
F = (−i + 2j + 3k )N
s = (0i + 0j + 4k )m
Work = F ٠ s = (−i + 2j + 3k) ٠(0i + 0j + 4k )
= 0 + 0 + 12
Work = 12 joule

37. 5. Prove that the vector A = i + 2j + 3k and B = 2i − j are
perpendicular to each other.
A = i + 2j + 3k
B = 2i − j + 0k
A ٠ B = 0
A ٠ B = (i + 2j + 3k) ٠(2i − j + 0k)
= 2 - 2 + 0
= 0
A and B 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟

38. 6. Find the value of n, so that the vectors A = 2i + nj + k
and B = 4i − 2j −2k are perpendicular to each other.
A = 2i + nj + k
B = 4i − 2j −2k
A ٠ B = 0
A ٠ B = (2i + nj + k) ٠(4i − 2j − 2k)
0= 8 – 2n + 2
2n =6
n= 3

39. Vector or Cross Product of Two Vectors
The vector product of two vectors is equal to the product of their magnitudes and the sine
of the smaller angle between them.
It is denoted by x (cross).
𝑨 x 𝑩 = AB sin θ n
sin θ =
𝑨 x 𝑩
𝐴𝐵
n =
𝑨 x 𝑩
|𝑨 x 𝑩|
The direction of unit vector n can be obtained from right hand thumb rule.
If fingers of right hand are curled from A to B through smaller angle
between them, then thumb will represent the direction of vector (A x B).

40. Properties of Vector Product
(i) Vector product is not commutative, i.e.,
𝑨 x 𝑩 ≠ 𝑩 x 𝑨
(𝑨 x 𝑩 ) = - (𝑩 x 𝑨 )
(ii) Vector product is distributive,
𝑨 x (𝑩 + 𝑪 ) = (𝑨 x 𝑩) + (𝑨 x 𝑪 )
(iii) Vector product of two parallel vectors is zero, i.e.,
𝑨 x 𝑩 = AB sin O° = 𝟎

41. (iv) Vector product of any vector with itself is zero.
𝑨 x 𝑨 = AA sin O° = 𝟎
(v) The vector or cross product of two vectors is also a vector.
Physical quantities such as
Torque  = 𝒓 x 𝑭
Angular momentum 𝑳 = 𝒓 x 𝒑

42. 𝑨 = 𝑨 𝒙 𝒊 + 𝑨 𝒚 𝒋 + 𝑨 𝒛 𝒌
𝑩 = 𝑩 𝒙 𝒊 + 𝑩 𝒚 𝒋 + 𝑩 𝒛 𝒌
𝑨 x 𝑩 =
𝒊 𝒋 𝒌
𝑨 𝒙 𝑨 𝒚 𝑨 𝒛
𝑩 𝒙 𝑩 𝒚 𝑩 𝒛
= 𝒊( 𝑨 𝒚 𝑩 𝒛 − 𝑨 𝒛 𝑩 𝒚) - 𝒋( 𝑨 𝒙 𝑩 𝒛 − 𝑨 𝒛 𝑩 𝒙) + 𝒌( 𝑨 𝒙 𝑩 𝒚 - 𝑨 𝒚 𝑩 𝒙)
= 𝒊( 𝑨 𝒚 𝑩 𝒛 − 𝑨 𝒛 𝑩 𝒚) + 𝒋( 𝑨 𝒛 𝑩 𝒙 − 𝑨 𝒙 𝑩 𝒛 ) + 𝒌( 𝑨 𝒙 𝑩 𝒚 - 𝑨 𝒚 𝑩 𝒙)

43. 1. If 𝑨 = 𝒊 + 𝟑 𝒋 + 2 𝒌 𝑎𝑛𝑑 𝑩 = 𝟑 𝒊 + 𝒋 + 2 𝒌 find vector product.
𝑨 = 𝒊 + 𝟑 𝒋 + 2 𝒌
𝑩 = 𝟑 𝒊 + 𝒋 + 2 𝒌
𝑨 x 𝑩 =
𝒊 𝒋 𝒌
𝟏 𝟑 𝟐
𝟑 𝟏 2
= 𝒊(3x2−2x1) - 𝒋(1x2 − 2x3) + 𝒌(1x1-3x3)
= 𝒊(6−2) - 𝒋(2 − 6) + 𝒌(1-9)
= 4 𝒊 + 4 𝒋 - 8 𝒌
= 4( 𝒊 + 𝒋 - 2 𝒌)

44. 2. Prove that the vectors 𝑨 = 𝟐 𝒊 − 𝟑 𝒋 − 𝒌 𝑎𝑛𝑑 𝑩 = −𝟔 𝒊 + 𝟗 𝒋 + 3 𝒌 are parallel.
𝑨 = 𝟐 𝒊 − 𝟑 𝒋 − 𝒌
𝑩 = −𝟔 𝒊 + 𝟗 𝒋 + 3 𝒌
𝑨 x 𝑩 = 𝟎
𝑨 x 𝑩 =
𝒊 𝒋 𝒌
𝟐 −𝟑 −𝟏
−𝟔 𝟗 3
= 𝒊[(-3x3)−(−1x9)] - 𝒋[(2x3) − (−1x − 6)] + 𝒌[(2x9)-(-3x-6)]
= 𝒊(-9+9) - 𝒋(6 − 6) + 𝒌(18-18)
= 0 𝒊 + 0 𝒋 - 0 𝒌
= 𝟎

45. 3. Determine the unit vector perpendicular to both 𝑨 = 𝟐 𝒊 + 𝒋 + 𝒌 𝑎𝑛𝑑
𝑩 = 𝒊 − 𝒋 + 2 𝒌 .
𝑨 = 𝟐 𝒊 + 𝒋 + 𝒌
𝑩 = 𝒊 − 𝒋 + 2 𝒌
n =
𝑨 x 𝑩
|𝑨 x 𝑩|
𝑨 x 𝑩 =
𝒊 𝒋 𝒌
𝟐 𝟏 𝟏
𝟏 −𝟏 2
= 𝒊[(1x2)−(1x − 1)] - 𝒋[(2x2) − (1x1)] + 𝒌[(2x-1)-(1x-1)]
= 𝒊(2+1) - 𝒋(4 − 1) + 𝒌(-2-1)
= 3 𝒊 - 3 𝒋 - 3 𝒌

46. 𝑨 x 𝑩 = 3 𝒊 - 3 𝒋 - 3 𝒌
| 𝑨 x 𝑩 |= 32 + (−3)2+(−3)2
= 9 + 9 + 9
= 27
= 3 3
n =
𝑨 x 𝑩
|𝑨 x 𝑩|
n =
3𝒊 − 3𝒋 − 3 𝒌
3 3
=
𝒊 − 𝒋 − 𝒌
3

47. In a straight line are:
𝒗 = 𝒖 + 𝒂𝒕
𝒔 = 𝒖𝒕 +
𝟏
𝟐
𝒂𝒕 𝟐
𝒗 𝟐
= 𝒖 𝟐
+ 𝟐𝒂𝒔
In a plane are:
𝑟2 − 𝑟1 = 𝑣0 𝑡 +
1
2
𝑎𝑡2
𝑣 = 𝑣0 + 𝑎𝑡 𝑣 𝑦 = 𝑣0𝑦 + 𝑎 𝑦 𝑡𝑣 𝑥 = 𝑣0𝑥 + 𝑎 𝑥 𝑡
𝑦 − 𝑦0 = 𝑣0𝑦 𝑡 +
1
2
𝑎 𝑦 𝑡2
𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 +
1
2
𝑎 𝑥 𝑡2
𝑣 𝑦
2 = 𝑣0𝑦
2 + 2𝑎 𝑦(𝑦 − 𝑦0)
𝑣 𝑥
2
= 𝑣0𝑥
2
+ 2𝑎 𝑥(𝑥 − 𝑥0)
𝑣 2 = 𝑣0
2 + 2𝑎 ( 𝑟2 − 𝑟1)
The equations of motion

48. Projectile Motion
A body thrown with some initial velocity and then allowed to
move under the action of gravity alone, is known as a
projectile.
If we observe the path of the projectile, we find that the
projectile moves in a path, which can be considered as a
part of parabola. Such a motion is known as projectile
motion.

49. Examples of projectiles are
(i) a object thrown from an aeroplane
(ii)a javelin or a shot-put thrown by an athlete
(iii)motion of a ball hit by a cricket bat
(iv)an arrow released from bow, etc.
Assumptions of Projectile Motion:
(1) There is no resistance due to air.
(2) The effect due to curvature of earth is negligible.
(3) The effect due to rotation of earth is negligible.
(4) For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in
magnitude and direction.

50. Angle of projection:
The angle between the initial direction of projection and the horizontal
direction through the point of projection is called the angle of projection.
Velocity of projection:
The velocity with which the body is projected is known as velocity of
projection.

51. Trajectory:
The path described by the projectile is called the trajectory.
Range:
Range of a projectile is the horizontal distance between the point of
projection and the point where the projectile hits the ground.
Time of flight:
Time of flight is the total time taken by the projectile from the instant of
projection till it strikes the ground.
Maximum height reached by the projectile:
The maximum vertical displacement produced by the projectile is known
as the maximum height reached by the projectile.

52. Types of Projectile Motion
Projectile fired at an angle with the horizontal (oblique projection):
For maximum range Rmax
sin 2θ = 1, (i.e) θ = 45°
Therefore the range is maximum when the angle of projection
is 45°.

53. Projectile fired parallel to horizontal (Horizontal projectile motion):