Bagdi 2 & 3 Dimension Motion 1 Bagdi

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 Bagdi 2 & 3 Dimension Motion 2 Bagdi

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Need of Vectors: -A particle confined to a straight line (one dimensional motion) can move in only two directions,
either upward or downward from the selected origin or toward or right or left from the origin. We can take the
motion of the particle to be positive in one of these directions and negative in the other direction. Therefore, in one-
dimensional motion direction is self-defined.
However, in two or three-dimensional motion magnitude and direction both are required to completely define it..
For example, a body is displaced through 10 m then there are infinite possible position i.e not completely defined. If
we specify direction also then it gives exact meaning e.g. 10 m from east to north.
Physical quantities can be divided into two classes viz.
(i) Scalars:- The physical quantities that possess magnitude only and no direction are called scalars e.g., mass,
distance, time, temperature etc. For example, when we say that the mass of a body is 2 kg, we have the complete
information. To ask its direction is a meaningless question. Scalars can be added, subtracted multiplied or divided
by using the rules of ordinary algebra.
(ii)Vectors:- The physical quantities that possess magnitude as well as direction are called vectors e.g.,
displacement, velocity, acceleration, force etc. For example, when we say that the displacement of a particle is 5 m,
the description is incomplete because direction of the displacement is not given. However, the displacement of a
particle is 5 m toward east is correct and meaningful.
Denotation of a Vector: - a vector is represented by bold letter, or by placing an arrow above it. The vector

representing a velocity would therefore appear as ‘v’, or by placing an arrow above it v .
 
The magnitude of a vector, a scalar quantity, is called modulus of the vector. The modulus of the vector OP = A , is

denoted by A or A.
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Graphical Representation of Vectors: -a vector is represented by a straight line that is drawn in the same direction
as the vector and has a length proportional to the magnitude of the vector. An arrow-head indicates directions of the
vector. The end where the arrow head is marked is called the head or tip of the vector, whereas the other end is
called the tail of the vector.
S.N. SCALAR QUANTITIES VECTOR QUANTITIES
1 Magnitude only
+ Both magnitude and direction +

2 Independent of direction+ Length represents magnitude, arrow shows direction +

3 Examples of scalars time, speed, volume, Examples of vectors force, velocity, and
and temperature acceleration
Different types of vectors
(i) Unit vector: -A unit vector of the given vector is a vector of unit magnitude and has the same direction as that
of the given vector. It is unit less and dimensionless vector and represents direction only.


    
A
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Unit vector of A is A  .In Cartesian coordinates, i , j , k are the unit vectors along x, y and z-axis.
A
(ii) Equal Vectors: -Two vectors are said to be equal if they have equal magnitude and same direction.
(iii) Negative Vector: - A negative vector of a given vector is a vector of same magnitude but acting in a direction
opposite to that of the given vector.
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(iv) Co-initial vectors: - The vectors are said to be co- initial, if their initial point is common.
(v) Collinear vectors: - These are those vectors, which are having equal or unequal magnitudes and are acting
along the parallel straight lines.
(vi) Coplanar Vectors: - those vectors, which are acting in the same plane, are called Coplanar Vectors.
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(vii) Null vector or zero vectors: -A vector having zero magnitude and no specific direction (arbitrary direction) is

called a null vector or zero vector. It is represented by a dot and denoted by O .
Examples: -1.The position vector of a particle at the origin is a zero vector. 2. The displacement vector of
stationary object is a zero vector. 3. The acceleration vector of a particle moving with uniform velocity is a zero
vector.
  
Properties: -1)If any vector is added to zero vector we get the same vector. A  O  A

2.When a real number is multiplied with zero vector we get zero vector. n  O  0 , n-real number.
  
3. When a vector is multiplied with zero vector we get zero vector A  O  O
  
4) If a zero vector is subtracted to any vector we get the same vector. A  O  A
(viii) Position vector: - A vector which represent position of an object in the co-
 
ordinate system is called position vector. OA  r is called position vector of the
object at point A and is generally represented by.
    
Position vector of object at A(x, y, z) is , r = OA = x i + y j + z k
(ix) Displacement vector is that vector which tells how much and in which direction an object has changed its
position in a given interval of time.
At t = 0 sec an object is at A at time t it displaced to B.
   
If the coordinates of point A is (x, y), then position vector of object at A is , r = OA = x i + y j

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   
If the coordinates of point A B is (x/, y/ ) , then position vector of object at B is , r = OB = x/ i + y/ j
  
Displacement vector, AB = S =(x/— x) i + (y/ — y) j . y
(x) Orthogonal unit vectors ˆi , ˆj kˆ
and are called orthogonal unit vectors. These vectors ˆj
must form a Right Handed Triad (It is a coordinate system such that when we Curl the
k̂ x
fingers of right hand from xto y then we must get the direction of z along thumb). The î
Axial vector z
ˆi  x , ˆj  y , kˆ  z  x  xˆi , y  yˆj , z  zkˆ
Axis of rotation
x y z
(xi) Polar vectors : These have starting point or point of application .
Example displacement and force etc. Anticlock wise rotation
Clock wise rotation
(xii) Axial Vectors :These represent rotational effects and are always along the Axis of rotation Axial vector
axis of rotation in accordance with right hand screw rule. Angular velocity, torque and angular momentum, etc., are
example of physical quantities of this type.
Addition of Vector:
(1)Triangle law of Vector addition: - statement-“if two vectors acting on a particle are represented in
magnitude and direction by the two sides of a triangle taken in one order their resultant vector is represented in
magnitude and directed by the third side of the triangle taken in opposite order”.
Draw equal

vector of P
and put tail of

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Q vector on

head of P .
  
To find their resultant R ,.Then the single vector drawn from the tail of P to the head of Q gives the resultant
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  
vector R . P and Q are represented by the sides OA and AC of a triangle OAC, taken in the same order. The third

side OC of the triangle taken in the opposite
opposi order represents their resultant R .
  
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OA  AC  OC Hence P  Q  R
(2) Parallelogram law of vectors: - It states that-““ if two vectors are represented in magnitude and direction
by the two adjacent sides of a parallelogram then their resultant vector is represented in magnitude and direction by
the diagonal of parallelogram drawn from the same point.
Let two vectors and be represented in magnitude and direction by the adjacent sides and of the
parallelogram OACB (Fig). Suppose the angle between the vectors is AOB= . According to parallelogram law
of vector addition, the diagonal represents the resultant (= ) in magnitude and direction. Suppose
make angle  with i.e., AOB= .
Magnitude of the resultant: - From C, draw perpendicular CD on the base
OB, meeting OB at point D,, when extended. Then, from right angled triangle
AEC, we have OC2 = OD2 + CD2 = (OA + AD)2+ CD2
OC2= OA2 + 2. OA.AD + AD2 + CD2
Now in triangle CBE, BE2 + CE2 = BC2
AC =OA + 2OA.AD + CD2………....(1)
2 2

AD
From right angledADC, cos   AD = AC cosAD = Q
AC
cos…....(2) Putting value in eq1.R2 = P2 + Q2 + 2 P. Q cos.
Magnitude of the resultant R  P 2  Q 2  2 P Q cos 

Direction of the resultant: -Suppose that the resultant R makes an angle  with the direction of P . Then, from
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CD CD
right angled triangle AEC, we have tan    - - - -(3)
OD OA  AD
CD
angled triangle BEC, we have sin  
From right-angled CD = AC sin  CD = Q sin …....(4)
AC
Q sin  1  Q sin  
Putting value ineq3from Eq4 &eq2. tan      tan  
P  Q cos   P  Q cos  
Thus, the magnitude of the resultant is given by eq (iii)
( and the direction of with is given by eq. (iv).
Different cases:- (a) When the vectors act along the same direction i.e.,  = 0.
R  P  Q  Hence R = P + Q
2
R P 2  Q 2  2 P Q cos 0  P 2  Q 2  2 P Q
Q sin 0
tan    0 Therefore, the magnitude of the resultant in this case is the sum
P  Q cos 0
 
of the magnitudes of two vectors and the direction of the resultant is along the direction of P & Q .
(b)) When the vectors are at right angles i.e.,  = 90, R P 2  Q 2  2 P Q cos 90  P 2
 Q2 
Q sin 90 Q 1 Q
tan    Hence   tan  
P  Q cos 90 P P
(c) When the vectors act along opposite direction i.e.,  = 180 , R  P 2  Q 2  2 P Q cos 180
R  P2  Q2  2 P Q R   P  Q 2 Hence R = P  Q
Thus the magnitude of the resultant in this case is equal to the difference in the magnitudes of the vectors.
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Q sin 0
tan    0 The resultant acts in the direction of the larger vector.
P  Q cos 0
(3) Polygon law of vectors addition: -It states that –‘if
number of vectors, acting at a point are represented in
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magnitude and direction by various sides of an open polygon

taken in the same order their resultant is represented in
magnitude and direction by the closing side of the polygon
taken in opposite order”.
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Polygon law of vectors is an extension of the triangle law of

vectors.
Let us illustrate the tail-to-tip
tip method of vector addition considering four vector . Applying tail-to-
tip method, these vectors are represented in magnitude and direction by the sides of the
regular polygon represented as shown in.. Then, the resultant is represented in magnitude and direction forms the
tail of the first vector to the tip of the last vector. In other words, the closing side of the polygon taken
ta in the
opposite order represents the resultant in magnitude and direction
i.e., = ., = = + + +
 
Subtraction of Vectors: - In order to subtract vector B from vector A first reverse the direction of
   
B thus production  B Then add A and  B by parallelogram law of vector addition.
 
Consider two vectors A and B of the same kind and inclined to each other at an angle as shown in Fig. (i).
 
(i) In order to find A  B reverse the
 
direction of B thus producing  B as
shown in Fig. (ii).
 
Then find the sum of A and  B by
parallelogram law of vector addition.

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The required difference is and is shown in Fig.
(ii) To find reverse the direction of thus producing as shown in Fig. (iii).
Resolution of a vector: -Splitting a vector in two or three parts in different direction such that the sum
(resultant) of all these parts is vector itself.
We generally resolve a vector into two components at right angles to each other. These are called rectangular
components of the
he vector. Consider a vector in the plane. Ax and Ay are ( rectangular components)
x-component and y-component
component respectively of vector along X-axis and Y-axis .
     
Hence from triangle law of vector addition OQ  OP  PQ OR A  Ax  A y
  
A  i Ax  j A y - - -(1)
In triangleOPQ cos = OP/OQ OP = OQcosHence
OQcos Ax  A cos - - - (2)
In triangleOPQ sin  = QP/OQ QP = OQ sin Hence Ay  A sin  - - - (3)
Magnitude:-Squaring and adding eq2
&3.

Ay A sin 
Direction:-Dividing eq3 by4 
Ax A cos 
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Ay  Ay 
tan   So   tan
1
  is angle
ngle between with
Ax  Ax 
-axis,
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Rectangular Components in Three Dimensions:-So So far we have considered the rectangular components of a
vector in a plane. Treatment can be extended to three-dimensional
three dimensional space in which case a vector will have three

rectangular components. Consider a vector in space. It has three rectangular components and as
       
shown in Fig. A  Ax  A y  Az A  i Ax  j A y  k Az - -
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-(1)

Magnitude of A A  Ax2  A y2  Az2

Note that is the

-component of is -component of and is the -component of
Relative velocity: General Formula: The relative velocity of a particle P1

moving with velocity v1 with respect to another particle P2 moving with v1
    v2
velocity v 2 is given by, v r12 = v1 – v 2 P2
(i) If both the particles are moving in the same direction then :  r12  1 –  2
P1
(ii) If the two particles are moving in the opposite direction, then :  r12  1   2
(iii) If the two particles are moving in the mutually perpendicular directions, then:  r12  12   22
(iv) If the angle between

1 and  2 be


, then  r12  12   22 – 21 2 cos  1/ 2
.

(a) Relative velocity of rain :If rain is falling vertically with a velocity vR  – vM 
vR vR

and an observer is moving horizontally with speed vM the velocity of rain   vR 
vM vM
  
relative to observer will be v RM  v R  v M
Fig. 0.17
hich by law of vector addition has magnitude v RM  v R2  v M2
Which
direction   tan 1 (v M / v R ) with the vertical as shown in fig.
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
(b) Relative velocity of swimmer : If a man can swim relative to water with velocity v and water is flowing
    
relative to ground with velocity vR velocity of man relative to ground vM will be given by: v  v M  v R , i.e.,
  
vM  v  vR
So if the swimming is in the direction of flow of water, v M  v  v R
And if the swimming is opposite to the flow of water, v M  v  v R
r
(c) Crossing the river: Suppose, the river is flowing with velocity  r . A man can swim in still water with velocity
r
m . He is standing on one bank of the river and wants to cross the river, two cases arise.
(i) To cross the river over shortest distance : That is to cross the river straight, the man should

swim making angle 
A vr B
with the upstream as shown.
    
Here OAB is the triangle of vectors, in which OA  v m , AB   r .  
vr
  w  v
Their resultant is given by OB   . The direction of swimming makes angle vm

 with upstream. From the triangle OBA,, we find,

r 
cos   Also sin   r Upstream O Downstream
m m
Where  is the angle made by the direction of swimming with the shortest distance (OB (OB) across the river.
Time taken to cross the river :If w be the width of the river, then time taken to cross the river will be given by
w w
t1  
  m2 –  r2
(ii) To cross the river in shortest possible time : The man should swim perpendicular to the bank.
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w
The time taken to cross the river will be: t 2 
m

A vr B
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 
w vr
vm
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Upstream O Downstream
In this case, the man will touch the opposite bank at a distance ABdown
down stream. This distance will be given by
w 
AB   r t 2   r or AB  r w
m m

Multiplication ofa Vector by a Scalar: -The multiplication of a vector A by a real number ‘n’ becomes

another vector n A . Its magnitude becomes n times the magnitude of the given vector Its direction is the same or
 
opposite as that of A, according as n is a positive or negative real number. n  A = n A .
Multiplication of Vectors: - It is of two kinds of multiplication operations for vectors:
(i) Scalar product or dot product of vectors:-
vectors Multiplication of one vector by a second vector so as to
produce a scalar. It is called scalar product or dot product of vectors.
Consider two vectors and with angle between them.
The Scalar product of two vectors and is defined as the product of
magnitude of one vector (say A) and the scalar component of the other vector B

cos along the direction of the first vector A .
   
A  B  A  component of B along A  OP  OS - - --(1)
 
In triangle OSQ 
OS = OQ cos= Bcos putting value in eq1. A  B  AB cos  - - - (2)

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Since A, B and are scalars, the product (read “ dot ”) is called scalar product, dot symbol is used
for multiplication so also called dot product.. Each of the vector and has a direction but the scalar product
 
itself does not have a direction. e.g. Work done, , Power =F v

Properties of Scalar Product:(i) For given vectors and B the value of the scalar product depends upon the
angle between them
For
For
For
Thus the dot product of two mutually perpendicular
vectors is zero.
(ii) The dot product of two vectors obeys commutative law. This directly follows from the definition of dot
 
product. A  B  AB cos  - - - -(1)
(1)
   
B  A  BA cos(  )  BA cos  Hence B  A  AB cos  - - - (2)
   
From eq1 & 2, B  A  A  B
This simply means that the order of vectors in the dot product does not matter.
 
(iii) Unit Vectors and the Cross Product:-- i  i  1  1 cos 0  1
     
Similarly i  i  j  j  k k  1
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       
i  j  1  1  cos 90  0 Similarly i  j  j k  k i  0
(iv) Consider two three-dimensional
dimensional vectors and These can be expressed in the rectangular form as :
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With the distribution law, it will yield,

Also
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Thus we can find the angle between the vectors and . Note that and are the magnitudes of vectors

and respectively.
(ii)Vector (Cross)Product:- Multiplication of one vector by a second vector so as to produce another vector. It
is called vector product or cross product of vectors.
Consider two vectors and where the angle is the angles between the two vectors.
The Vector product of two vectors and is defined as the product of magnitude of one vector (say A) and the
 
component of the other vector B sin perpendicular to the direction of the first vector A & n .
     
A  B  A  component of B  to A  n  OP  QS  n - - - (1)
  
In triangle OSQ QS = OQ sin= B sin putting value in eq1. A  B  AB sin  n - - - (2)
The vector or cross product of vectors and is another
vector symbolized by; due to this
It is called vector product of vector.
(i) The magnitude ofvector C is given by;
Here & are the magnitudes of vectors & while
is the smaller of the angles between vectors and

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(ii) is perpendicular to the plane containing and and its direction is given by right-hand
right rule.Curl the
fingers of the right hand in the sense that fore finger represent first vector & middle finger represent second

vector through the smaller angle between them. The extended thumb gives the direction of n or i.e. resultant.
  
Properties of vector product:-(i) We know that A  B  AB sin  n - - - - -(1)
   
B  A  BA sin(  ) n   AB sin  n - - - - (2)
Thus referring to Fig.,, the direction of vector is opposite
to that of vector .
The cross product of two vectors does not obey commutative
law i.e.,
Since the magnitude in each case is it follows, therefore, that .

(ii) Suppose two vectors and are parallel or antiparallel. The angle between them is either or
  
Then A  B  AB sin 0  n Hence The cross product of parallel (or antiparallel) vectors is zero.
(iii) The cross product of a vector with itself is zero i.e.
(iv) The cross product obeys distributive law i.e.,
(v) The magnitude of the cross product of two vectors is equal to the area of parallelogram formed by them.
Suppose two vectors and are represented in magnitude and direction
by the two adjacent sides and of the parallelogram OLKM.
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The area of the parallelogram = OLMNMN

= B ( A sin ) = AB sin
But is the magnitude of the cross product Therefore, the
magnitude of the cross product of two vectors is equal to the area of the
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parallelogram formed by them.

(vi) Unit Vectors and the Cross Product:--The
The cross product of two vectors can be evaluated in terms of the
rectangular components. To do so, we must consider the cross product of the unit vectors.
        
(a) Twoequal vectors are parallel . i  i  1  1  sin 0 n  0 Similarly, i i  j  j  k  k  0
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    
(b) i  j  1  1  sin  n  n  k
  
The direction of i  j i.e. n is given by the right-hand
right hand rule and it is along the positive Z-axis.
Z But this is just the
     
unit vector Therefore we have i  j  k . Simply reversing the order of the unit vectors gives j  i   k
           
Similarly, j  k i &k  j  i and k  i j &i  k  j
Consider three dimensional vectors and These vectors can be expressed in terms of rectangular vectors as:
       
A  i Ax  j Ay  k Az & B  i B x  j B y  k B z
  
i j k
       
A  B  ( i Ax  j Ay  k Az )  ( i B x  j B y  k B z )  Ax Ay Az
Bx By Bz

Motion of an Object in Plane (2D motion):-

motion): in case of two-
dimensional motion, the particle moves in a plane. Suppose a particle follows a
curved path in the XY plane as shown in Fig.. At time t1, the particle is at point
P1 (x1, y1) & at time t2, it is at point P2 (xx2, y2). The position of the particle in the
plane can be located in two ways.
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   
Position vector of object at P1is , r 1 = OA = x1 i + y1 j
   
Then position vector of object at P2is , r 2 = OB = x2 i + y 2 j
in time interval, t = t2 – t1 the
The Displacement Vector:- as the particle goes from point P1 to point P2,in
Displacement vector,
    
S = r 2 r 1 =(x2— x1) i + (y 2 — y1 ) j .
  
 r =x i + y j - - - - -- (1)
  
According to triangle law of vector addition, r 2 = r 1 +r
 

Displacement vector,  r = r 2  r1
The Velocity Vector: - We define the velocity of the particle as the displacement vector divided by the time
  
r r 2  r1 
interval t. the direction of velocity is the same as that of displacement. i.e., v   ……...(2)
t t
 
 x i  y j   x  y  x y
v  vx i  vy j  i  j  vx  & vy 
t t t t t
are velocity along X & Y direction.
(i) The magnitude of the velocity is v= v x2  v 2y . (ii) The direction in which the particle is heading at any time
vy
may be described in terms of angle  between velocity vector and X-axis as : tan  .
BA

vx
Instantaneous acceleration vector:- The instantaneous acceleration vector is defined as the limit of the average
    
 v v 2  v1  vx  vy 
 approaches zero i.e., a 
acceleration vector as the time interval t  
t t t
GD

  v x  v y v x v y
ax i ay j  i  Hence acceleration along X & Y direction is a x  & ay 
t t t t

Equations of motion:-uniform velocity v of the particle is given by ;
I

    
 Displacement r r 2  r1  
(x - x ) i + (y 2 - y1 ) j
v   vx i  vy j  2 1
time t t t
   
v x t i  v y t j  (x2 - x1 ) i + (y 2 - y1 ) j Comparing coefficient of x & y component of both side:
vx t = x 2  x1 x2 = x1 + v x t ...(ii)
& vy t = y2  y1 y2 = y1 + vy t ...(iii)
1stEquations of motion:-Then uniform (constant) acceleration of the particle is given by ;
 
 Final velocity  Initial velocity v  u      
a   at v u Hence v  u  a t
time taken t
     
v x i  v y j  (u x i + u y j)  (a x t i + a y t j) Compare the component along X & Y direction.
vx = ux + ax t ...(ii) This is Ist eq. of motion along X-axis.
X
st
and v y = u y + ay t ...(iii) This is I eq. of motion along Y-axis.
Y
2nd Equation of Motion:-Since acceleration is constant, the average velocity during the time interval t is
 
 Final velocity  Initial velocity v  u
But v av= 
2 2


   

 v u  (u  a )  u 
Displacement = Average velocity  time S( )t    t
2  2
 

Bagdi 2 & 3 Dimension Motion 10 Bagdi

 Bagdi 2 & 3 Dimension Motion 11 Bagdi
  1     1  1 
S  ut  at Hence (x2 - x1 ) i + (y 2 - y1 ) j  (u x t i + u y t j)  ( a x t 2 i + a y t 2 j)
2 2 2
 1 1
x 2 - x1  u x t i + a x t 2 & y 2 - y1  u y t  ayt 2
2 2
Hence 2nd eq. of motion along X & Y direction are
1 1
x 2  x1  u x t + a x t 2 & y 2  y1  u y t  ayt 2
2 2
3 Equation of motion:- v x  u x  2 a x S x
rd 2 2
& v y  u 2y  2 a y S y .
2

PROJECTILE MOTION:-If an object is thrown into air at any angle with vertical (  0)then motion of the
object under the influence of gravity is called projectile motion& the body thrown is called projectile. The path of
the projectile is parabolic called trajectory.
Examples are thrown base ball, a speeding bullet and an athlete doing the high jump. Since the projectile moves
horizontally as well as vertically, it is a case of two dimensional motion.
Oblique Projectile Motion:-Suppose a projectile is launched from the origin O(0,0) with an initial velocity u
at an angle  above the horizontal as shown in Fig.. At t = 0, the projectile is at the origin O so that x0 = 0 and y0 =
0. Suppose at any time t, the projectile is at point P(x, y). This means that at time t, the projectile has covered a
horizontal distance x from the origin O and attained a height y above the origin.

At time t = 0, from resolution of vector the x and y components of u are ux = ucos and uy = u sin  respectively.
1) Motion along the horizontal:- The projectile has motion along the horizontal (X-axis) at constant velocity
ux= ucos because there is no horizontal acceleration (since ay = 0) so applying 2nd equation of motion along x-
1 1
direction. S x  u x t + a x t 2 or x  0  u cos  t + 0  t 2
2 2
BA

x = u cos t Hence t = x/ u cos - - - -(1)

2) Motion along the vertical;- At the same time motion along the vertical (Y-axis) at constant acceleration
(ay=g). Applying 2nd equation of motion along y-direction.
1 1
S y  u yt + ayt 2 or y  0  u sin   t + g  t 2 - - -(2) From eq1 & 2.
GD

2 2
x 1 x sin  g
y  u sin   + g ( )2 y x +  x2
u cos  2 u cos  cos  2 u cos 
2 2

2
y = ax + ax . This is eq of motion for a projectile. It is similar to eq of parabola, hence path projectile is parabolic.
I

g
Here a = tan& b = are the constants.
2 u cos 2 
2

Velocity of Projectile:-After time ‘t’ the horizontal components vx of the velocity is constant (i.e., ax = 0).
vx = ucos - - - -(i)
At time t, the vertical component of its velocity at this time is given by ; vy = uy – g t
or vy = u sin  – g t ...(iii)
At time t, velocity v  v 2x  v 2y  (u cos  ) 2  (u sin   gt ) 2
Parameters of projectile motion:- The important parameters of projectile motion are time of flight (T), the
maximum height (H) attained and the horizontal range (R).
(i) Time of flight (T):-The time for which the projectile remain in air before coming to ground is called time of
flight. OR It is the time elapsing from the launching to the time the projectile returns to the ground again.
1
Applying 2nd equation of motion along y-direction. S y  u y t + a y t 2
2
When projectile returns to ground, t = T, its vertical displacement is
1
zero (i.e., y = 0). 0  0   u sin   T + g  T 2
2
(usin – ½ g T)T= 0
T = 0 OR usin – ½ g T = 0 usin = ½ g T hence
2 u sin 
T  - - - -(1)
g
Bagdi 2 & 3 Dimension Motion 11 Bagdi
 Bagdi 2 & 3 Dimension Motion 12 Bagdi
Note that there are two solutions for t. The first solution (t = 0) corresponds to the launching point (origin O). The
second solution [T= (2u sin )/g] corresponds to the point when the projectile returns to the ground (point D).
Note. In Fig. When the projectile is launched upward at an angle with the horizontal, its vertical velocity decreases
with time and becomes zero at the highest point (point B) of the projectile’s path. At this instant, the projectile starts
downward journey. The time (tm) taken by the projectile to reach maximum height can be found by setting
vy = 0, in 1steq. of motion along Y-axis vy = uy + ay t
u sin  T
0 = u sin  – g tm u sin  = g tm t m   - - - -(2)
g 2
Hence, time taken by the projectile to reach maximum height is equal to one-half of the total time of flight (T). This
symmetry is due to the fact that the acceleration due to gravity (g) is the same for upward and downward motion.
(ii) Maximum height attained (H). It is the maximum height (vertical displacement) to which the projectile rises
above the launching point (origin). Applying 3rdeq of motion along Y-axis, v 2y  u 2y  2 a y S y
At maximum height velocity along Y-axis is vy= 0. 0  (u sin  ) 2  2 g H
u 2 sin 2 
H - - - - -(3)
2g
(iii) Horizontal Range (R). It is the horizontal displacementtravelled by the projectile before returning to the
ground (i.e., original height). In other words, it is the horizontal displacementtravelled by the projectile during time
1
T (= time of flight). Applying 2nd equation of motion along y-direction. S x  u x t + a x t 2
2
1 2 u sin  2 sin  cos 
R  u cos  T + 0  T 2  u cos   OR R  u 2 
2 g g
u 2 sin 2
BA

R ( sin 2 = 2 sin cos)

g
Note that the horizontal range (R) of a projectile depends upon the projection angle () and initial speed (u).
u 2 sin 2
Special Case :-(a)Maximum Range:-, the range will be maximum when R  = maximum,
GD

g
For a given initial speed, it is only possible when , sin 2 = 1 (maximum)
u2
it is only possible when 2 = 90 0 or  = 45 0. Hence Maximum range Rmax 
g
I

Hence for a given initial speed, the horizontal range of the projectile will be maximum when the angle of projection
is 450 (neglecting air friction). for small projection angles (< 450), the horizontal range is small. It is also small
when the projection angles are large (> 450).
(b) For the given initial speed u, there are two projection angles  and (900 – ) for
which the horizontal range of the projectile is the same.
u 2 sin 2
For projection angle , the horizontal range is R  - - - - (1)
g
For projection angle (900– ), the horizontal range is
u sin 2(90   )
2
u sin (180  2 )
2
u sin 2
2
R  OR R 
g g g
This is just the same equation as eq. (iv). Although projection angles  and
(900 – ) give the same horizontal range (for same initial speed u).The two paths
differ in time of flight (T) and maximum height (H) attained because the values of T and H depend upon the angle
of projection ().
Horizontal Projectile:- Suppose at t = 0 a body is thrown horizontally (i.e., parallel to earth’s surface) from

a tower with a velocity u as shown in Fig. At this instant, downward vertical velocity uy = 0.If it hit the ground
after t sec. The motion of the body can be split into two parts :
(i) Vertically downward motion with constant acceleration (ay = g). The downward velocity of the ball increases in
magnitude by 9.8 m s–1 in each second. At point A, the ball has no vertical speed, at point B (after 1s), its vertical
speed is 9.8 m s–1; at C (after 2s) 19.6 m s–1 and so on.
Bagdi 2 & 3 Dimension Motion 12 Bagdi
 Bagdi 2 & 3 Dimension Motion 13 Bagdi
Applying 2nd equation of motion along y-direction.
1 1
S y  u yt + ayt 2  H  0  t + gt 2
2 2
(2 H ) / g  t 2

2H
Time taken to reach the ground is t  - - - -(1)
g

(ii) Horizontal motion with constant velocity u i.e. Acceleration
along x-axis ax = 0. The horizontal velocity of the body remains the
same till it hits the ground. It is because there is no horizontal
component of acceleration; a x  0 .
Applying 2nd equation of motion along x-direction.
1 1
S x  u x t + a x t 2 or R  u  t + 0  t 2
2 2
2H
From eq1 R  u  From the above discussion, the following conclusions can be drawn :
g
(a) The projectile motion is a two dimensional motion with constant acceleration. (b) During projectile motion,
the horizontal velocity vx remains the same but vertical velocity changes at a constant rate with time.
Thus in above case, the magnitude of downward vertical velocity (vy) increases by 9.8 m s–1 after every second.
(c) The time of flight of the projectile depends upon the downward motion alone.
(d) The velocity of the ball changes continuously in both magnitude and direction. If the vertical component of
velocity at any time is vy= vy + ay t then magnitude of velocity of the ball is v = v 2x  v 2y at that time.
BA

CIRCULAR MOTION:-If a body moves along a circular path with uniform speed then motion of the body is
called uniform circular motion.The word “uniform” refers to the constant speed.
e.g., rotation of pulleys, wheels, shafts,the orbits of the planets and their natural satellites are
nearly circular etc.
GD

Although the magnitude of the velocity remains constant in this case, In a circular motion,
the direction of motion is tangential to the circular path so direction of velocity is changing
continuously. Since acceleration is defined as the rate of change of velocity, a change in
direction of velocity constitutes acceleration just as does a change in magnitude. Thus a
I

particle undergoing uniform circular motion is accelerating.

Since the path of the particle is circular, the direction of velocity vector will be at right angles
to the radius of the circle at every point on the path. So linear acceleration act along the
radius towards centre so called centripetal acceleration.
(i) Angular displacement. The angle turned by a particle moving along a circular path in a given time is called the
angular displacement of the particle during that time interval.
Consider a particle moving along a circular path with uniform speed ‘v’ around centreO. Let us consider O as the
origin of our coordinate system. Suppose initially (i.e., at t = 0), the particle is at point P. At time t1, the particle is at
point P1 and its angular position is 1. At time t2, the particle is at point P2 and its angular position is 2. During the
time interval t2 – t1 (= t), the angular displacement is 2 – 1=. The S.I. unit of angular displacement is radian.
(ii) Average angular velocity (). It is the angular displacement traveled by a
body in unit time interval. If the angular displacement of a particle is  during

the time interval t, then average angular velocity of the particle is  
t

 2 r
 OR    2 v
t T
The S.I. unit of average angular velocity is radians/second (rad/s).
(iii) Time period. The time taken by the particle to complete one revolution is
called time period of the circular motion. It is denoted by T.
(iv) Relation between linear speed and angular speed:-Consider a particle moving with constant speed v (linear)
in a circle of radius r as shown in Fig. Since linear speed of the particle ‘v’ is constant, its angular speed  will also
Bagdi 2 & 3 Dimension Motion 13 Bagdi
 Bagdi 2 & 3 Dimension Motion 14 Bagdi
ppose in a small time interval t, the particle moves from point A to point B and covers a distance
be constant. Suppose
lalong
along the circumference of the circle. Note that during this time interval, the angular
displacement of the particle is .
Arc l
Angle     - - - - (1)
Radius r
 l   
Dividing equation by t. r  Hence v   r v    r
t t
i.e., Linear speed = Angular speed × Radius of circular path
Linear speed v (= r)) will be greater for particles farther away from the axis than for particles nearer to the axis.
(v) Expression for linear acceleration:-- Consider a particle moving with a constant speed v round a circle of

radius r as shown in Fig. At point A,, its velocity v A is in the direction of the tangent to the path at point A. After a
short time t,, the particle is at point B and its

velocity v B is in the direction of the tangent to the
path at point B.. During the time interval t, the
particle moves from A to B, covering a small
distance l along the arc which subtends a small
angle  at centre O.. Magnitude of velocity at points
A and B are the same (since speed is constant).
However, directions of velocity differ at points A and
B.. Hence the particle has acceleration. The change in
  
velocity vector v  v B  v A

BA

Δv
The acceleration of the particle is a  - - - (2)
t
(a) Direction of acceleration:- In the figure, we notice that when t is very small (approaching zero), l and 
  
are also very small. Consequently, v B will be almost parallel to v A and v will be essentially perpendicular to
GD

  
them. Thus v is along the radius of the circle. Since a , by definition, is in the same direction as v , the
acceleration must point towards the centre of the circle. Therefore, this acceleration is called centripetal

acceleration (centripetal means “centre seeking”) and
and henceforth we shall denote it by a c.
I

(b) Magnitude of centripetal acceleration (a( c):- In fig-(i) OAB

Arc l
Angle     - - (1)
Radius r
  
The vector v B, v A and v in Fig. 9.16 (ii) form a triangle that is geometrically similar to the ABO in
   
Fig(i).Angle between v B& v A is & v B  v A  v , speed of the body.
Arc v
Thus Angle     - -(ii) Comparing eq1 & 2.
Radius v
v l v v l v v2
  Dividing eq by t ,   v ac 
v r t r t r r
It is Centripetal acceleration,
Thus a particle moving in a circle of radius r with a constant speed v has an acceleration directed toward the centre
of the circle of magnitude ac = v2/r = 2 r is called centripetal acceleration. Since v and r are constant, the
magnitude of centripetal acceleration is constant.
However, the direction of the acceleration is not constant; it is changing continuously
ntinuously as
 
shown in Fig. Note that v and a care perpendicular to each other at every instant.
(i) ac = v2/r. ac depends upon v and r.. It is because the greater the speed v,, the faster the
velocity changes direction and the larger the radius, the less rapidly the velocity changes
direction.

Bagdi 2 & 3 Dimension Motion 14 Bagdi

 Bagdi 2 & 3 Dimension Motion 15 Bagdi
(ii) The velocity of the particle is always tangential to the circle and centripetal acceleration always acts radially
toward the centre. For this reason, centripetal acceleration is sometimes called radial acceleration.
(iii) If the particle is accelerating toward the centre, why does it remain a constant distance away rather than
“falling” into the centre. In fact, if the particle did not accelerate, it would move straight ahead and leave its circular
path around the centre. Because of its centripetal acceleration, the particle does, in a sense, fall toward the centre,
but while it falls, it continues to move forward. The combination of the two motions keeps the particle on a circular
path of constant radius.
(vi) Centripetal force: - Due to centripetal acceleration force on body F = mv2/r = m2r must act towards the
centre is called centripetal force.e.g. Gravitation force of attraction between sun & a planet provide the necessary
centripetal force, electric force of attraction between nucleus & an electron provide the necessary centripetal force,
friction force between road &tyre provide the necessary centripetal force.
(vii) Centrifugal Force:-According to Newton’s 3rd law of motion every action has equal but opposite reaction so a
force F = mv2/r = m2r along radially outward direction must act on the body called Centrifugal Force.
VECTORS
TYPE-A: Based on Composition of Vectors
Q. 1 ABCDE is a pentagon. Prove that AB  BC  CD  DE  EA  0 .
Q. 2 A particle has a displacement of 12 m towards east and 5 m towards the north and then 6 m vertically upward.
Find the magnitude of the sum of these displacements. Ans = 14.32 m
Q. 3 Two vectors, both equal in magnitude, have their resultant equal in magnitude of the either. Find the angle
between the two vectors. Ans120
Q. 4 The resultant vector of P and Q is R . On reversing the direction of Q , the resultant vector becomes S . Show
that : R2 + S2 = 2 (P2 + Q2)
Q. 5 Calculate the angle between a 2N force and a 3N force so that their resultant is 4N. (Ans  = 7531).
BA

Q. 6 The greatest and the least resultant of two forces acting at a point are 29 N and 5 N respectively. If each force
is increased by 3N, find the resultant of two new forces acting at right angle to each other. (Ans25 N,  = 3652).
Q. 7 Establish following vector inequalities: (i) | a  b || a |  | b | (ii) | a  b || a || b | When does equality sign
apply?
GD

Q. 8 Establish following vector inequalities: (i) | a – b || a |  | b | (ii) | a – b || a | – | b | When does equality sign
apply?
Q. 9 On a certain day, the rain was falling vertically with a speed of 30 ms–1. A wind started blowing, with a speed
of 10 ms–1 from the north to south, find the direction in which boy should hold his umbrella in order to protect him
I

from the rain. (Ans. 1826)

Q. 10 A river 800 m wide flows at the rate of 5 Kmh–1. A swimmer who can swim at 10 kmh–1 in still water ,
wishes to cross the river straight. (i) Along what direction must he strike? (ii) What should be his resultant velocity?
(iii) How much time he would take ? Ans 30, 2.4 m/s, 333.3 s.
Q. 11 Two forces whose magnitudes are in the ratio of 3:5 give a resultant of 35 N. If the angle of inclination be
60, Calculate the magnitude of each force. (Ans. 15N, 25N)
Q. 12 Two equal forces have the square of their resultant equal to three times their product. Find the angle between
them. (Ans. 60)
Q. 13 At what angle do the two forces (P + Q) and (P – Q) act so that the resultant is 3P 2  Q 2 .(Ans. 60)
Q. 14 The resultant of two equal forces acting at right angles to each other is 1414 dyne. Find the magnitude of
either force. (Ans.1000 dyne)
TYPE-B: Based on Relative Velocity of two inclined Motions
Q. 1 Rain is falling vertically with a speed of 30 ms–1. A woman rides a bicycle with a speed of 10 ms–1 in the north
to south direction. What is the relative velocity of rain with respect to the woman? What Is the relative velocity of
rain with respect to the woman? What is the direction in which she should hold her umbrella to protect herself from
the rain?(Ans.31.6m/s, 1826 )
Q. 2 In a harbour, wind is blowing at the speed of 72 kmh–1 and the flag on the mast of a boat anchored in the
harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 kmh–1 to the north , what is the
direction of the flag on the mast of the boat? (Ans. 45.01 i.e. east)
Q. 3 To a man walking due east at the rate of 2 kmh–1 , rain appears to fall vertically. When he increases his speed
to 4 kmh–1 it appears to meet him at an angle of 45 . Find the real direction and speed of rain. (Ans. 45)

Bagdi 2 & 3 Dimension Motion 15 Bagdi

 Bagdi 2 & 3 Dimension Motion 16 Bagdi
Q. 4 A train is moving with a velocity of 30 Km h–1 due east and a car is moving with a velocity of 40 Km h–1 due
north. What is the velocity of car as appears to a passenger in the train ? (Ans. 30 Km h–1, 3652 West of North)
TYPE-C: Based on Expressing the vectors in terms of Base Vectors
Q. 1 Find unit vector parallel to resultant of the vectors A  iˆ  4 ˆj  2kˆ and B  3iˆ – 5 ˆj  kˆ .(Ans. [ 4iˆ – ˆj  kˆ ] /32)
Q. 2 Find unit vector parallel to resultant of vectors A  2iˆ – 6 ˆj – 3kˆ and B  4iˆ  3 ˆj – kˆ [Ans. (6iˆ – 3 ˆj – 4kˆ) / 61 ]
Q. 3 Find the value of  in the unit vector 0.4iˆ  0.8 ˆj  kˆ [Ans. 0.2 ]
TYPE-D: Based on Rectangular components of vectors
Q. 1 A force is inclined at 50 to the horizontal . If its rectangular component in the horizontal direction be 50 N,
find the magnitude of the force and its vertical component. Ans. 77.78N, 59.58 N,
Q. 2 A child pulls a rope attached to a stone with a force of 80 N. The rope makes an angle of 40 to the ground
(i) Calculate the effective value of the pull tending to move the stone along the ground
(ii) Calculate the force tending to lift the stone. [Ans. (i) 45.96 N (ii) 38.57 N]
TYPE-E: Based on Scalar or Dot Product of two vectors
Q. 1 Find the angle between the vectors A  iˆ  2 ˆj – kˆ and B  iˆ  ˆj – 2kˆ [Ans. 60]
Q. 2 Prove that the vectors A  iˆ  2 ˆj  3kˆ and B  2iˆ – ˆj are perpendicular to each other.
Q. 3 Find value of  so that vectors A  2iˆ  ˆj  kˆ & B  4iˆ – 2 ˆj – 2kˆ are perpendicular to each other.[Ans.  = 3]
Q. 4 A body acted upon by a force of 50 N is displaced through a distance of 10 m in a direction making an angle
of 60 with the force. Calculate the work done by the force. [Ans. 250 J]
Q. 5 Three vectors A, B and C are such that A  B  C and their magnitudes are 5, 4, and 3 respectively. Find the
angle between A and C [Ans. 53]
Q. 6 If | A  B || A – B | find the angle between A and B . [Ans. 90]
BA

Q. 7 If iˆ & ĵ are unit vectors along X & Y-axis respectively, then what is the magnitude & direction of iˆ  ˆj &
iˆ – ˆj ? [Ans. 45, 45]

Q. 8 If unit vectors â and b̂ are inclined at angle  , then prove that | a – b | 2Sin
GD

2
Q. 9 If A  B  C and A + B = C , then prove that A and B are perpendicular to each other.
2 2 2

Q. 10 The sum and difference of two vectors A and B are A  B  2iˆ  6 ˆj  kˆ and A – B  4iˆ  2 ˆj – 11kˆ . Find the
I

magnitude of each vector and their scalar product A.B [Ans. 50 , 41,–25 ]
Q.11 If A , B & C have magnitudes 8, 15 & 17 units & A  B  C , find the angle between A & B .[Ans. 90]
 A2  B 2 – C 2 
Q. 12 If A  B – C , then determine the angle between A and B .  Ans.  Cos –1 
 2 AB 
Q. 13 Prove that ( A  2 B).(2 A – 3B)  2 A 2 ABCos – 6 B 2 .
TYPE-F: Based on vectors or Cross Product of two Vectors
Q. 1 Prove that the vectors A  2iˆ – 3 ˆj – kˆ and B  –6iˆ  9 ˆj  3kˆ are parallel.
Q. 2 Calculate the area of the parallelogram whose two adjacent sides are formed by the vectors A  3iˆ  4 ˆj and
B  3iˆ  7 ˆj [Ans. 33 m2]
Q. 3 Determine a unit vector perpendicular to both A  2iˆ  ˆj  kˆ and B  iˆ – ˆj  2kˆ [Ans. (iˆ  ˆj  kˆ ) / 3 ]
Q. 4 Determine the sine of the angle between the vectors 3iˆ  ˆj  2kˆ and 2iˆ – 2 ˆj  4kˆ . [Ans 2/7]
Q. 5    
Show that A – B  A  B  2 A  B 
Q. 6    
For any three vectors A, B and C , prove that A  B  C  B  C  A  C  A  B  0  
Q. 7  2
For any two vectors A and B , prove that A  B  A 2 B 2 – A.B   2

Q. 8 Find | A B | if | A | 10, | B | 2 and A.B  12 [ Ans 16]

Bagdi 2 & 3 Dimension Motion 16 Bagdi

 Bagdi 2 & 3 Dimension Motion 17 Bagdi
ˆ ˆ ˆ ˆ ˆ ˆ
Q. 9 The diagonals of a parallelogram are given by the vectors 3i  j  2k and i – 3 j  4k . Find the area of the
parallelogram. [Ans. 8.66 Sq. Units]
  
Q.10 If A  2iˆ  3 ˆj  kˆ and B  3iˆ  2 ˆj  4kˆ ,then find the value of A  B  A – B [Ans. – 20iˆ  10 ˆj  10kˆ ]
Q.11 Find the value of  for which the vectors 3iˆ  3 ˆj  9kˆ and iˆ  ˆj  3kˆ are parallel. [Ans. =2/3]


Q.12 Find a unit vector perpendicular to A  4i – Jˆ  3kˆ & B  – 2 iˆ  ˆj – 2 kˆ .[Ans. – iˆ  2 ˆj  2kˆ ]
1
2

ˆ
Q.13 If A & B are two such vectors that | A | 2 | B | 7 & A.  B  3iˆ  2 ˆj  6k find the angle between A & B .
[Ans.  / 6]
Q.14 Prove that | a  b |  a 2
b 2
– a . b 
2

MOTION IN TWO DIMENSIONS

TYPE-A: Based on Motion in a plane : Q. 4 A bullet fired at an angle of 30 with horizontal
Q. 1 On a ground, a motorist follows a track that turns hits ground 3 Km away. By adjusting angle of
to his left by an angle of 60 after every 500 m. Starting projection, can one hope to hit a target 5 Km away?
from a given turn, specify displacement of motorist at Assume the muzzle speed to be fixed and neglect air
third, sixth and eighth turn. Compare magnitude of resistance. [ Ans 3460m ]
displacement with the total path length covered by the Q. 5 A projectile has arrange of 50m and reaches a
motorist in each case. maximum height of 10 m. Calculate the angle at which
[Ans (i) 1000m, 1500m (ii) 3000m (iii) 866m] the projectile is fired. [ Ans38.66 ]
TYPE-B: Based on Projectile Fired Horizontally : Q.6 A boy stands at 39.2 m from a building and throws
Q. 1 A projectile is fired horizontally with a velocity of 98 a ball which just passes through a window 19.6m above
BA

ms–1 from the top of a hill 490 m high. Find (i) the time the ground. Calculate the velocity of projection of the
taken to reach the ground (ii) the distance of the target ball. [ Ans27.72 m/s ]
from the hill and (iii) velocity with which the projectile Q.7 Show that there are two angles of projection for
hits the ground. [Ans 10s, 980m, 138.59m/s 45 ] which the horizontal range is the same. Also show that
Q. 2 A bomb is dropped from an aeroplane when it is the sum of the maximum heights for these two angles is
GD

directly above a target a height of 1000m. The independent of the angle of projection.
aeroplane is moving horizontally with a speed of 500 Q. 8 A fighter plane flying horizontally at an altitude of
kmh–1. By how much distance will the bomb miss the 1.5 km with a speed 720 kmh–1 passes directly overhead
target ? [Ans1984m] an antiaircraft gun. At what angle from vertical should
I

Q. 3 Two tall buildings face each other and are at a the gun be fired if shell muzzle speed 600 ms–1 to hit
distance of 180 m from each other. With what velocity the plane? At what maximum altitude should the pilot
must a ball be thrown horizontally from a window 55 m fly the plane to avoid being hit? [Ans70.5, 16km ]
above the ground in one building , so that it enters a Q. 9 A hunter aims his gun and fires a bullet directly at a
window 10.9 m above the ground in the second monkey on a tree. At the instant the bullet leaves the
building ? [Ans 60 m/s ] barrel of the gun, the monkey drops. Will the bullet hit the
Q. 4 A marksman wishes to hit a target just in the same monkey? Give your answer with proper reasoning. [yes]
level as the line of sight. How high from the target he Q.10 A bomber, flying upwards at an angle of 53 with
should aim, if the distance of the target is 1600 m and the the vertical, releases a bomb at an altitude of 800 m. The
muzzle velocity of the gun is 800 ms–1? [Ans=19.6 m]. bomb strikes the ground 20 s after its release. Find (i) the
TYPE-C : Based on Projectile Fired at an angle velocity of the bomber at the time of release of the bomb,
with the Horizontal : (ii) the maximum height attained by the bomb, (iii) the
Q. 1 A body is projected with a velocity of 30 ms–1 at an horizontal distance travelled by the bomb before it strikes
angle of 30 with the vertical. Find the maximum height, the ground, (iv) the velocity (magnitude and direction) of
time of flight and the horizontal range. [ Ans 79.53m ] the bomb just when it strikes the ground. Take Sin 53 =
Q. 2 A cricketer can throw a ball to a maximum 0.8 , Cos 53 = 0.6, g = 10 ms–2. [ Ans 100m/s980m,
horizontal distance of 100 m. How high above the 1600m, 161.3 m/s 2945/ ]
ground can the ground can the cricketer throw the same Q.11 At what angle should a body be projected with a
ball ? [ Ans 50m ] velocity 24 ms -1just to pass over the obstacle 16 m high
Q. 3 The ceiling of a long hall is 25 m high. What is the at a horizontal distance of 32 m ? Take g =10 ms -2.
maximum horizontal distance that a ball thrown with a [ Ans 6754/ or 4840/ ]
speed of 40 ms–1 can go without hitting the ceiling of TYPE-D: Based on Uniform Circular Motion :
the hall ? [ Ans150.7m ]
Bagdi 2 & 3 Dimension Motion 17 Bagdi
 Bagdi 2 & 3 Dimension Motion 18 Bagdi
Q.1What is the angular velocity of a second hand and time period and (iv) centripetal acceleration. [ Ans 5
minute hand of a clock?[ Ans 0.1047rad/s, 0.0017rad/s ] rad/s, 0.8 Hz, 1.25 s, 50 m/s2 ]
Q. 2 Calculate the angular speed of fly wheel making Q. 4 An insect trapped in a circular groove of radius 12
420 revolutions per minute. [ Ans 44 rad/s ] cm moves along the groove steadily and completes 7
Q. 3 A body of mass 0.4 whirled in a horizontal circle revolutions in 100s. (i) What is the angular speed and
–2
radius 2m with a constant speed of 10 ms . Calculate the linear speed of the motion?(ii) Is the acceleration
its (i) angular speed (ii) frequency of revolution (iii) vector a constant vector? What is its magnitude? [ Ans
991.2 cm/s2 ]

Horizontal Projectile Motion 1 1

(a) tan 1   (b) tan   (c) tan 1 (1) (d) tan 1 (5)
1. An aeroplane is flying horizontally with a velocity of 5 5
600 km/h at a height of 1960 m. When it is vertically at a 10.A large number of bullets are fired in all directions
point A on ground, a bomb is released from it. Bomb with same speed v . What is maximum area on ground
strikes the ground at point B. Distance AB is on which these bullets will spread
(a) 1200 m (b) 0.33 km (c) 3.33 km (d) 33 km v2 v4 v4 v2
(a)  (b)  (c)  2 (d)  2
2. A ball is rolled off the edge of a horizontal table at a g g 2
g 2
g2
speed of 4 m/s. It hits the ground after 0.4 second. Oblique Projectile Motion
Which statement given below is true 1. A ball is thrown upwards at an angle of 60o to the
(a) It hits ground at a horizontal distance 1.6 m from the horizontal. It falls on the ground at a distance of 90 m.
edge of table If the ball is thrown with the same initial velocity at an
(b) The speed with which it hits the ground is 4.0m/s (c) angle 30o, it will fall on the ground at a distance of
Height of the table is 0.8 m (a) 30 m (b) 60 m (c) 90 m (d) 120 m
(d) It hits the ground at an angle of 60o to the horizontal 2. Four bodies P, Q, R & S are projected with equal
3. An aeroplane flying 490 m above ground level at 100 velocities having angles of projection 15o, 30o, 45o & 60o with
m/s, releases a block. How far on ground will it strike the horizontal respectively. The body having shortest range
BA

(a) 0.1 km (b) 1 km (c) 2 km (d) None is

4. A body is thrown horizontally from the top of a tower of (a) P (b) Q (c) R (d) S
height 5 m. It touches the ground at a distance of 10 m from 3. For a projectile, the ratio of maximum height reached
the foot of the tower. The initial velocity of the body is to the square of flight time is (g = 10 ms–2)
GD

(a) 2.5 ms–1 (b) 5 ms–1 (c)10 ms–1 (d)20 ms–1 (a) 5 : 4 (b) 5 : 2 (c) 5 : 1 (d) 10 : 1
5. An aeroplane moving horizontally with a speed of 4. A stone projected with a velocity u at an angle  with
720 km/h drops a food pocket, while flying at a height of the horizontal reaches maximum height H1. When it is
396.9 m. Time taken by a food pocket to reach the projected with velocity u at an angle  / 2    with the
ground & its horizontal range is (Take g = 9.8 m/s2) horizontal, it reaches maximum height H2. The relation
I

(a) 3 s & 2000 m (b) 5 s & 500 m between the horizontal range R of the projectile, H1 & H2 is
(c) 8 s & 1500 m (d) 9 s & 1800 m (a) R  4 H 1 H 2 (b) R  4(H 1  H 2 )
6. A particle (A) is dropped from a height and another
H12
particle (B) is thrown in horizontal direction with speed (c) R  4(H 1  H 2 ) (d) R
H22
of 5 m/s from the same height. The correct statement is
(a) Both particles will reach at ground simultaneously 5. An object is projected with a velocity of 20 m/s
(b) Both particles will reach at ground with same speed making an angle of 45o with horizontal. Equation for the
(c) Particle (A) will reach at ground first with respect to trajectory is h = Ax – Bx2 where h is height, x is
particle (B) (d) Particle (B) will horizontal distance, A & B are constants. The ratio A : B is
reach at ground first with respect to particle (A) (a) 1 : 5 (b) 5 : 1 (c) 1 : 40 (d) 40 : 1
6. Which of the following sets of factors will affect the
7. A particle moves in a plane with constant
horizontal distance covered by an athlete in a long–jump event
acceleration in a direction different from the initial (a) Speed before he jumps and his weight
velocity. The path of the particle will be (b) The direction in which he leaps and the initial speed
(a) A straight line (b)An arc of a circle (c) The force with which he pushes the ground and his speed
(c) A parabola (d) An ellipse (d) None
8. At the height 80 m, an aeroplane is moving with 150 m/s. 7. A ball thrown by one player reaches the other in 2
A bomb is dropped from it so as to hit a target. At what sec. Maximum height attained by the ball above the
distance from the target should the bomb be dropped
point of projection will be about
(a) 605.3 m (b) 600 m (c) 80 m (d) 230 m
(a) 10 m (b) 7.5 m (c) 5 m (d) 2.5 m [Pb. PMT 2002]
9. A bomber plane moves horizontally with a speed of
8. In a projectile motion, velocity at maximum height is
500 m/s and a bomb released from it, strikes the ground
in 10 sec. Angle at which it strikes the ground will be] (a) u cos  (b) u cos (c) u sin  (d) None [AIEEE 2002]
2 2

Bagdi 2 & 3 Dimension Motion 18 Bagdi

 Bagdi 2 & 3 Dimension Motion 19 Bagdi
9. If two bodies are projected at 30o and 60o 20. The horizontal range of a projectile is 4 3 times its
respectively, with the same velocity, then[CBSE PMT 2000] maximum height. Its angle of projection will be
(a) Their ranges are same (b) Their heights are same (a) 45 o (b) 60 o (c) 90 o (d) 30 o
(c) Their times of flight are same (d) All of these 21. A ball is projected upwards from the top of tower
10. A body is thrown with a velocity of 9.8 m/s making an angle with a velocity 50 ms 1 making an angle 30 o with the
of 30o with the horizontal. It will hit the ground after a time
horizontal. The height of tower is 70 m. After how many
(a) 1.5 s (b) 1 s (c) 3 s (d) 2 s
11. The equation of motion of a projectile are given by x = 36
seconds from the instant of throwing will the ball reach
2
t metre and 2y = 96 t – 9.8 t metre. Angle of projection is the ground [DPMT 2004]
(a) 2 s (b) 5 s (c) 7 s (d) 9 s
(a) sin 1  4  (b) sin 1  3  (c) sin 1  4  (d) sin 1  3  22. Two bodies are thrown up at angles of 45o and 60o,
5 5 3 4 
12. For a given velocity, a projectile has same range R respectively, with the horizontal. If both bodies attain
for two angles of projection if t1 and t2 are the times of same vertical height, then the ratio of velocities with
flight in the two cases then [AIEEE 2004]
which these are thrown is [DPMT 2005]
2 2 3 3
(a) t1 t 2 R2 (b) t1 t 2  R (c) t1 t 2  1 (d) t1 t 2 
1
(a) (b) (c) (d)
R R2 3 3 2 2
13. A body of mass m is thrown upwards at an angle  23. At what point of a projectile motion acceleration and
with horizontal with velocity v. While rising up its velocity are perpendicular to each other [Orissa JEE 2005]
velocity after t seconds will be (a) At the point of projection (b) At the point of drop
(a) (v cos  )2  (v sin )2 (b) (v cos  v sin )2  gt (c) At the topmost point
(d) Any where in between the point of projection and topmost point
(c) v 2  g 2 t 2  (2 v sin  ) gt
(d) v 2  g 2 t 2  (2 v cos  ) gt 24. An object is projected at an angle of 45 with the
14. A cricketer can throw a ball to a maximum horizontal. The horizontal range and the maximum
horizontal distance of 100 m. With the same effort, he height reached will be in the ratio.
throws the ball vertically upwards. The maximum (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 [Kerala PET 2005]
BA

height attained by the ball is [UPSEAT 2002]

(a) 100 m (b) 80 m (c) 60 m (d) 50 m
15. A cricketer can throw a ball to a maximum
horizontal distance of 100 m. Speed with which he
throws the ball is (nearest integer)
GD

(a) 30 ms–1 (b) 42 ms–1 (c) 32 ms–1 (d) 35 ms–1

16. A ball is projected with velocity Vo at an angle of
elevation 30. Mark the correct statement [MP PMT 2004]
(a) Kinetic energy will be zero at the highest point of
I

the trajectory
(b) Vertical component of momentum will be conserved
(c) Horizontal component of momentum will be conserved
(d) Gravitational potential energy will be minimum at highest
point of the trajectory
17. Neglecting air resistance, time of flight of a
projectile is obtained by
(a) U vertical (b) U horizontal [J & K CET 2004]
(c) U  U vertical  U horizontal
2 2
(d) U  U (U 2 vertical  U 2 horizontal )1 / 2
18. A ball is thrown from a point with a speed v o at an
angle of projection  . From same point and at the same
instant a person starts running with a constant speed
v o / 2 to catch the ball. Will the person be able to catch
the ball? If yes, what should be the angle of projection
(a) Yes, 60 o (b) Yes, 30 o (c)No (d) Yes,
19. A stone is thrown at an angle  to the horizontal
reaches a maximum height H. Then the time of flight of
stone will be
2H 2H 2 2 H sin  2 H sin 
(a) (b) 2 (c) (d) [BCECE 04]
g g g g

Bagdi 2 & 3 Dimension Motion 19 Bagdi