Motion And Force

1

MOTION AND FORCE CHAPTER 3 1. Displacement 2. Velocity 3. Acceleration 4. Velocity-Time Graph 5. Equations of Uniformly Accelerated Motion 6. Newton’s Laws of Motion 7. Momentum 8. Elastic and Inelastic Collision 9. Force Due to Water Flow 10.Momentum and Explosive Forces 11.Rocket Propulsion 12.Projectile Motion Arshad Ali 4/25/2007

1

MOTION AND FORCE

The branch of mechanics which deals with relations of force, matter and motion is called kinematics. Motion may be defined as a continuous change of position. We live in a universe of continual motion. In every piece of matter, the atoms are in a state of never ending motion. We move around the earth’s surface, while the earth moves in its orbit around the sun. The sun and the starts, too, are in motion everything in the vastness of space is in a state of continuous motion.

1. Displacement

The shortest distance between two points is called displacement. OR The displacement is a change in the position of a body from its initial position to its final position. It is a vector quantity and its SI unit is meter m. The dimensions of displacement are [L].

Explanation

The displacement can be represented as a vector that describes how far and in what direction the body has been displaced from its original position. The tail of displacement vector is located at the position where the displacement started, and its tip is located at the final position where the displacement ended. For example, if a body is moving along a curve as shown in Fig. with A as its initial position and B as its final position, then the displacement d of the body is represented by AB. The displacement d occurs with a change in position. It is the difference between a final position vector and an initial position vector.

Fig. If r1 is the position vector of A and r2 is the position vector of point B, then by head to tail rule it can be seen from the Fig. that d r2 r1 When a body moves along a straight line, the displacement coincides with the path of motion as shown in Fig.

2. Velocity

Time rate of change of displacement is called velocity. OR The distance covered by a body in a unit time in a particular direction. If ∆d is the change in displacement in time ∆t, then velocity v is given by V

………. i

It is a vector quantity and its SI unit is ms . The dimension of velocity is [LT-1]. -1

Fig.

1

Average Velocity

If d is the total displacement covered by a body in time t, then its average velocity during the time interval t is written as Total Displacement Average Velocity = Total Time d v av = ………. ii t Average velocity does not tell us about the motion between two points. The path may be straight or curved and the motion may be steady or variable. For example if a squash ball comes back to its straight point after bouncing off the wall several times, its total displacement is zero and so also its average velocity.

Instantaneous Velocity

The velocity of a body at some one instant of time, or at any point of its path is called its instantaneous velocity.

Explanation

In order to understand the concept of instantaneous velocity, consider a body moving along a path ABC in xy-plane. At any time t, let the body be at point A as shown in Fig. Its position is given by position vector r1. After a short time interval ∆t following the instant t, the body reaches the point B which is described by the position vector r2. The displacement of the body during this short time interval is given by ∆d r2 r1

Fig. This instantaneous velocity at a point A can be found by making ∆t smaller and smaller. In this case ∆d will also become smaller and point B will approach A. if we continue this process, letting B approach A, thus allowing ∆t and ∆d to decrease but never disappear completely, the ratio ∆d ∆t approaches a definite limiting value which is the instantaneous velocity. Therefore Δd v ins = lim ………. iii Δt → 0 Δt If the instantaneous velocity does not change, the body is said to be moving with uniform velocity.

3. Acceleration

The time rate of change of velocity of a body is called acceleration. It is a vector quantity and its SI unit is ms-2. It has dimension [LT-2]. The acceleration of an object characterizes how rapidly its velocity is changing both in magnitude and direction.

Average Acceleration

Consider a body whose velocity v1 at any instant t changes to v2 in further small time interval ∆t. The two velocity vectors v1 and v2 and change in velocity ∆v v2 v1 are represented in Fig. The average acceleration aav during time interval ∆t is given by

1

a av =

v 2 − v1 Δv = Δt Δt

Fig.

Instantaneous Acceleration

Acceleration of a body at a particular instant is know as instantaneous acceleration and its vale is obtained from the average acceleration as ∆t is made smaller and smaller till it approaches zero. Mathematically; Δv a ins = lim Δt → 0 Δt

Positive Acceleration

If the velocity of a body is increasing then its acceleration will be negative.

Uniform Acceleration

If the velocity of a body changes by equal amount in equal intervals of time, the body is said to be have uniform acceleration. It should be noted that if a body is moving with uniform acceleration, its average acceleration is equal to instantaneous acceleration.

4. Velocity-Time Graph

A graph which is used to show the variation of velocity of an object with time is called velocity time graph. When the velocity of the car is constant, its velocity time graph is a horizontal line as shown in Fig.1

Fig. 1 When the car moves with constant acceleration, the velocity time graph is a straight line which roses the same height for equal intervals of time as shown in Fig.2

Fig. 2 The average acceleration of an object can be determined by finding the slope of its velocity-time graph. When the car moves with increasing acceleration, the velocity-time graph is a cure as shown in Fig.3 The magnitude of the instantaneous acceleration at this instant is numerically equal to the slope of the tangent at the point A on the velocity-time graph of the object as shown in Fig.3

1

Fig. 3 The distance moved by an object can also be determined by using its velocity-time graph. For example Fig.1 shows that the object moves at constant velocity v for time t. The distance covered by the object will be S = v × t ………. i This distance can also be found by calculating the area under velocity time graph. This are is shown shaded in Fig.1 and is equal to Area = v × t ………. ii Now consider the case of Fig.2, here velocity of the object increases uniformly form 0 to v in time t. The magnitude of its average velocity will be 0+v v av = 2 1 v av = v 2 So, the distance covered will be S = vav × t 1 S = v× t ………. iii 2 Now are of the triangle shaded in Fig.2 will be 1 Area = × base × height 2 1 Area = × v × t ………. iv 2 Considering the above two examples it is a general conclusion that the are between the velocity time graph and time axis is numerically equal to the distance covered by the object. 5.

Equations of Uniformly Accelerated Motion

Consider an object is moving with uniform accelerating ‘a’ along a straight line. Let vi be the initial velocity of the object which becomes vf after some time t and the object covers a distance S, and then we have v f = v i + at  v + vf  S = v av × t =  i × t  2  1 S = v i t + at 2 2 2 2aS = v f − v i2

6. Newton’s Laws of Motion Newton’s laws of motion are empirical laws which can be deduced from experiments. Newton published his laws in 1987 in his famous book called “Principia Mathematica”. Newton’s laws are valid for speeds that are low compared with the speed of light. For very fast moving objects, such as atomic particles in an accelerator, relativistic mechanics developed by Albert Einstein is applicable.

First Law of Motion

A body at rest will remain at rest, and a body moving with uniform velocity will continue to do so, unless acted upon by some unbalanced external force. This is also known as law of inertia.

Inertia

The property of an object tending to maintain the state of rest or state of uniform motion is referred to as the object’s inertia.

1

The more inertia, the stronger is this tendency in the presence of a force. Thus, The mass of the object is a quantitative measure of its inertia.

Inertial Frame of Reference

The frame of reference in which Newton’s first law of motion holds good is known as inertial frame of reference.

Non-Inertial Frame of Reference

The frame of reference in which Newton’s first law of motion does not hold good is known as non-inertial frame of reference.

Second Law of Motion

A force applied on a body produces acceleration in its own direction. The magnitude of acceleration produced varies directly with the applied force and inversely with the mass of the body. Let ‘m’ be the mass of a body and we apply a force ‘F’ on it, if ‘a’ is the acceleration produced in it, then we can write a ∝ F ………. i 1 a ∝ ………. ii m Combining i and ii we have F a∝ m F ∝ ma F = kma Letting k = 1 ………. iii F = ma This is mathematically form of Newton’s law. The SI unit of force is Newton and it is defined as

Newton

One Newton is the force which produces an acceleration of 1ms-2 in a body of mass 1kg. Thus 1N = 1kg×1ms-2.

Third Law of Motion

Action and reaction are equal in magnitude but opposite in direction. i.e.; whenever on body exerts a force on another, but in opposite direction for the same length of time and has the same line of action. Note: Action and reaction forces never act on the same body.

7. Momentum

The quantity of motion in a body is known as the momentum. OR The product of mass ‘m’ and velocity ‘v’ is known as momentum. Momentum is denoted by symbol P, we can write P = mv ………. i Momentum is a vector quantity. Its direction is same as that of velocity. The SI unit of momentum is kgms-1. It is also expressed as Ns. s Kgms-1 = kgms-1 × s = kgms-2 × s = Ns

Newton’s Second Law in term of Momentum

Newton’s second law of motion can also be stated in terms of momentum as follows: Time rate of change of momentum of a body equals the applied force. Consider a body of mass ‘m’ moving with an initial velocity vi. Let a force F acts upon it for time‘t’ after which velocity becomes vf. The acceleration ‘a’ produced by this force is given by;

1

a= By second’s law

vf − vi t a=

From i and ii

………. i

F ………. ii m

F vf − vi = m t F × t = m( v f − v i ) F × t = mvf − mvi ………. iii Where mvi is the initial momentum and mvf is the final momentum of the body. Eq.iii shows that change in momentum is equal to the product of force and the time for which force is applied. This form of second law is more general than F =ma, because it can easily be extended to account for changes as the body accelerates when its mass also changes. For example, as a rocket accelerates, it loses mass because its fuel is burnt and ejects to provide greater thrust. From eq.iii we can write mv f − mvi F= ………. iv t Thus time rate of change of momentum of a body equals to the applied force. This is Newton’s second law in term of momentum.

Impulse

When a very large force acts upon a body for a short interval of time and changes its momentum, the product of such force and time is called impulse. Impulse = F × t It is a vector quantity and its direction is same as that of P.

Impulsive Force

Large force acting for a short time is called an impulsive force.

Isolated System

The system in which two or more bodies exert force on one another and no external agency exerts any force on them is called an isolated system of interacting bodies. For example the molecules of a gas enclosed in a vessel at constant temperature form an isolated system of interacting bodies. The molecules can collide with one another because of their random motion but being enclosed by glass vessel, no external agency can exert a force on them.

Law of Conservation of Momentum

The total linear momentum of an isolated system remains constant provided no external forces act on the system. Consider an isolated system of two smooth hard interacting balls of masses m1 and m2, moving along same straight line, in the same direction with velocities v1 and v2 respectively. Both balls collide and after collision, ball of mass m1 moves with velocity v1' and the m2 with v '2 in the same direction as shown in Fig.

Fig. To find the change in momentum of mass m1, using the relation F × t = mvf − mvi

1 We have

F × t = m1 v1' − m1 v1 ………. i Similarly for the ball of mass m2, we have F'×t = m 2 v '2 − m 2 v 2 ………. ii Adding I and ii we have, ( F + F') t = ( m1v1' − m1v1 ) + ( m 2 v '2 − m 2 v 2 ) ………. iii Since the action of force F’ is equal and opposite to the reaction force F, we have F = -F’, so the L.H.S. of eq.iii becomes zero, hence 0 = ( m1v1' − m1v1 ) + ( m 2 v '2 − m 2 v 2 ) In other words [Change in momentum of 1st ball + Change in momentum of 2nd ball] = 0 m1v1 + m 2 v 2 = m1v1' + m 2 v '2 ……….. iv Or Which means that total initial momentum of the system before collision is equal to the total final momentum of the system after collision. Consequently, the total change in momentum of the isolated ball system is zero.

8. Collision

When two objects come close tot each other so that there is some soft of interaction between them, due to which they strike with one another, we say that collision has taken place between the objects.

Elastic Collision

The collision in which kinetic energy K.E. of the system remains constant is called elastic collision. For example, when a hard ball is dropped onto marble flour, it rebounds to very nearly the initial light. It looses negligible amount of energy in the collision with the floor. So in this case K.E. of the ball is conserved.

Inelastic Collision

The collision in which kinetic energy K.E. of the system does not remain constant is called inelastic collision. For example, when tow tennis balls collide then after collision they will rebound with velocities less that the velocities before impact. During this process, a portion of K.E. is lost, partly due to friction and partly due to its change into heat and sound energies.

Elastic Collision in One Dimension

Consider two smooth non-rotating balls of masses m1 and m moving initially with velocities v1 and v2 respectively, in the same direction. They move along the same straight line without rotation. Let their velocities after collision is v1' and v '2 as shown in Fig.

Fig. By applying the law of conservation of momentum, i.e, [Momentum before collision] = [Momentum after collision] m1 v1 + m 2 v 2 = m1 v1' + m 2 v '2

m1v1 − m1v1' = m 2 v 2 − m 2 v '2

………. i m1 ( v1 − v1' ) = m 2 ( v 2 − v '2 ) As the collision is elastic, so by applying law of conservation of energy, we have; [K.E. before collision] = [K.E. after collision]

1

1 1 1 1 m1v12 + m 2 v 22 = m1v1'2 + m 2 v '22 2 2 2 2 Multiplying both sides by 2, we have m1v12 + m 2 v 22 = m1v1'2 + m 2 v '22

m1v12 − m1v1'2 = m 2 v '22 − m 2 v 22

m1 ( v12 − v1'2 ) = m 2 ( v '22 − v 22 ) ……….. ii m1 ( v1 + v1' )( v1 − v1' ) = m 2 ( v '2 + v 2 )( v '2 − v 2 ) Dividing eq. ii by eq. i, we have m1 ( v1 + v1' )( v1 − v1' ) m 2 ( v '2 + v 2 )( v '2 − v 2 ) = m1 ( v1 − v1' ) m 2 ( v '2 − v 2 ) v1 + v1' = v '2 + v 2 ……….. iii v1 − v 2 = v '2 − v1' v1 − v 2 = −( v1' − v '2 ) We know that v1 − v 2 is the velocity of first ball relative to the second ball before collision. Similarly v '2 − v1' is the velocity of the second ball relative to the first ball after collision. It means that relative velocities before and after the collision has the same magnitude but are reversed after the collision. From eq.iii we can write, ………. iv v '2 = v1 − v 2 + v1' Now writing eq.i as m1v1 − m1v1' = m 2 v '2 − m 2 v 2 ……….v Putting the values of v '2 from eq. iv and eq. v, we have; m1v1' = m1v1 + m 2 v '2 − m 2 ( v1 − v 2 + v1' ) m1v1' = m1v1 + m 2 v 2 − m 2 v1 + m 2 v 2 − m 2 v1' v1' ( m1 + m 2 ) = v1 ( m1 − m 2 ) + 2m 2 v 2 v ( m − m 2 ) + 2m 2 v 2 v1' = 1 1 ( m1 + m 2 ) ( m − m 2 ) v + 2m 2 v v1' = 1 ( m1 + m 2 ) 1 ( m1 + m 2 ) 2 ……….vi

Similarly, we can find value of v '2 , which will be ( m − m1 ) v 2m1 v '2 = v1 + 2 ( m1 + m 2 ) 2 m1 + m 2

………. vii

The eq.vi and vii gives the values of v1' and v '2 in terms of known quantities m1, m2, v1 and v2.

Special Cases

There are some cases of special interest,

Case-I when m1 = m2

Putting m1 = m2 = m in eq.vi

v1' =

( m − m ) v + 2m v ( m + m) 1 ( m + m) 2

 2m  v1' = 0 +  v 2  2m  ………. a v1' = v 2 Now for v '2 from eq. vii, we have;

1

2m ( m − m) v v1 + m+m ( m + m) 2  2m  v '2 =   v1 + 0  2m  ………. B v '2 = v1

v '2 =

Fig. From eq.a and eq. b it is clear that when two bodies of identical masses collide, they simply exchange their velocities.

Case-II when m1 = m2 and v2 = 0

Putting m1 = m2 = m and v2 = 0 in eq.vi ( m − m ) v + 2m ( 0) v1' = ( m + m) 1 ( m + m) v1' = 0 ………. c And from eq. vii, we have; ( m − m ) ( 0) 2m v '2 = v1 + ( m + m) m+m  2m  v '2 =   v1 + ( 0 )  2m  ………. D v '2 = v1

Fig. From eq. c and d it is clear that the ball of mass m1 after collision will come to rest and m2 will take off with velocity that of m1.

Case-III when a light body collides with a massive body at rest, i.e., m2>>m1 and v2 = 0 Putting v2 = 0 and m2>>m1 i.e., m1 is negligibly small and v2 = 0 in eq. vi. ( 0 − m 2 ) v + 2m 2 ( 0) v1' = (0 + m2 ) 1 (0 + m2 ) ( − m2 ) v + 0 v1' = ( m2 ) 1

v1' = − v And from eq. vii we have; v '2 =

………. e

( m − 0) ( 0) 2( 0 ) v1 + 2 0 + m2 (0 + m2 ) v '2 = 0 ………… e

1

Fig. Form eq. e and f it is clear that m1 after collision will bounce back with same velocity while m2 will remain stationary.

Case-IV when a massive body with light stationary body i.e., m1>>m2 and v2 = 0 Putting m1>>m2 i.e., m2 is negligibly small and v2 = 0 in eq. vi, we have; ( m − 0) v + 2( 0) ( 0) v1' = 1 ( m1 + 0 ) 1 ( m1 + 0 )

v1' = v1 And from eq. vii, we have; v '2 =

………. g

( 0 − m1 ) ( 0 ) 2m 1 v1 + m1 + 0 ( m1 + 0 )

v '2 = 2v1

……….. h

Fig. From eq. g and h it is clear that after the collision, there is no change in the velocity of massive body, but the lighter body bounces off in the forward direction with twice the velocity of the incident body.

9. Force due to Water Flow

When water from horizontal pipe strikes a wall normally, a force is exerted on the wall. Suppose the water strikes the wall normally with velocity v and comes to rest on striking the wall. Then, Change in velocity = Final velocity – Initial Velocity =0–v Change in Velocity = -v From Newton’s second law in term of momentum, the force is equal to change in momentum per seconds of water. If mass ‘m’ of the water strikes the wall in time t then force F extorted on the water is given as; F = Mass per second × Change in velocity m F = × ( − v) t v F = −m ………. 1 t From third law of motion, the reaction force exerted by the water on the wall is equal but opposite , hence;  v F = − m   t

1

v ………. 2 t Thus force can be calculated from the product of mass of water striking normally per second and change in velocity. Suppose the water flows through a pipe at 3kgs-1 and its velocity changes from 5ms-1 to 0 on striking the wall, then F = Mass × Change in velocity = 3kgs-1 × 5ms-1 F = 15N F=m

10. Momentum

and Explosive Forces

There are many examples where momentum changes are produces by explosive forces within an isolated system. For example, when a shell explodes in mid-air, are fragments fly off in different direction. The total momentum of all its fragments equals the initial momentum of the sell. Suppose a falling bomb explodes into two pieces as shown in Fig. The momentum of the addition is equal to the original momentum of the falling bomb.

Fig. Consider another example of bullet of mass ‘m’ fired from a rifle of mass M with a velocity v initially, the total momentum of the bullet and rifle is zero. From the principle of conservation of linear momentum, when the bullet is fired, the total momentum of bullet and rifle still remains zero, since no external force has acted on them. If v’ is the velocity of the rifle then Mv + Mv’ = 0 (bullet) (rifle) ⇒ Mv’ = -mv m v' = − v ……….. 1 M The momentum of the rifle is thus equal and opposite to that of the bullet. Since mass of rifle is much greater than the bullet, it follows that the rifle moves back or recoil with only a fraction of the velocity of the bullet.

11.Rocket Propulsion

A rocket is propelled by the ejection of a potion of its mass (burning gas) to the rear. The forward force on the rocket is the reaction to the backward force on the ejected material, and as more material is ejected, the mass of the rocket decreases, and the rocket is expelled with extremely high velocity.

1

Fig. The rocket gains momentum equal to the momentum of the gas expelled from the engine but in opposite direction. the rocket engines continue to expel gases after the rocket has begun moving and hence rocket continues to gain more and more momentum. So instead of travelling at steady speed the rocket gets faster and faster so longer the engines are operating. In order to provide enough upward thrust t o overcome gravity, a typical rocket consumes about 10,000kgs-1 of fuel and ejects the burnt gases at speeds of over 4000ms-1. In effect more than 80% of the launch mass of a rocket consists of fuel only. One way to overcome the problem of mass of fuel is to make the rocket from several rockets linked together. When one rocket has done its job, it is discarded leaving others to carry the space craft further up at even greater speed. If ‘m’ is the mass of the gases ejected per second with velocity v relative to the rocket, the change in momentum per second of the ejecting gases is mv. This equals the thrust produced by the engine on the body of the rocket. So, the acceleration ‘a’ of the rocket is mv a= M Where M is the mass of the rocket. When the fuel in the rocket is being consumed, mass of the rocket Ma continuously decreases and hence the acceleration increases until all fuel is burned.

12.Projectile

Any object that is given an initial velocity and it follows a path determined by the gravitational force acting on it and by the friction resistance of the atmosphere is called a projectile. For example, i. Missile shot from a gun. ii. A rocket after its fuel is exhausted. iii. A thrown cricket ball. iv. An objet dropped from an aero plane. v. A football thrown in air.

Projectile Motion

The two dimensional motion under constant acceleration due to gravity is called projectile motion. OR A motion of a body which has both vertical and horizontal components of motion is called projectile motion.

Trajectory

The path followed by the projectile is called its trajectory

Horizontal and Vertical Components of Distance

Suppose the ball leaves the hand of the thrower at point A as shown in Fig.

1

Fig. The velocity of the ball at that instant is completely horizontal. Let this velocity be vx. Since the only force on a projectile in this case is its weight. The x-component of the force on the projectile is then zero and the y-component is the weight of the projectile, i.e., ax = 0 and ay = g. As ax = 0, so there will be no acceleration in the horizontal direction. Thus, horizontal velocity will remain unchanged and will be vx, until the ball hits something. Hence horizontal distance x is given by; x = vx t ………. 1 Now vertical component will be; 1 y = viy t + a y t 2 2 Since initial velocity of the projectile along the vertical direction is zero i.e., v i y = 0 and as ay = g, so we can write 1 y = gt 2 ………. 2 2

Expression for Instantaneous Velocity of a Projectile

Suppose that a projectile is fired at an angle θ with initial velocity vi as shown in Fig.

Fig. The components of its initial velocity along horizontal and vertical direction are; v i x = v i cosθ ………. 3 v i y = v i sinθ ………. 4 As we know that the horizontal component of velocity remains constant, therefore, we can write; v f x = v i x = v i cosθ ………. 5 Note the vertical component of the velocity will be; vf y = vi y + a y t The initial vertical component v i y = v i sinθ in the upward direction, as the motion of the body is in the upward direction so ay = -g, so we can write v i y = v i sinθ − gt ………. 6 The magnitude of velocity at any instant is v = v f2 x + v f2 y

………. 7

Direction of Velocity ‘v’ Let φ be the angle which this resultant velocity makes with the horizontal can be found as

1

tanφ =  vf φ = tan -1  y  vf  x

vf y vfx

   

………. 8

Height of the Projectile

The maximum height which the projectile attains can be determined as; 2as = v f2 − v i2 As body moves upward, so a = -g, the initial component of vertical velocity viy = vi sinθ and final component vfy = 0, because the body comes to rest after reaching the highest point and s = h, therefore above equation becomes; - 2gh = 0 − v i2sin 2θ h=

v i2sin 2θ 2g

………. 9

Time of Flight

The time taken by body to cover the distance fro the place of its projection to the place where it hits the ground is called time of flight. The vertical distance covered by the body will be zero, because the body goes up and comes back to same level, thus S = h = 0. If the body is projecting with velocity v making angle θ with horizontal, then 1 its vertical component of velocity will be vi sinθ. Hence the equation S = v i t + gt 2 2 becomes 1 0 = v i sinθi − gt 2 2 2 v sinθ t= i ………. 10 g

Range of the Projectile

The range of projectile is the maximum distance which a projectile covers in the horizontal direction. To determine the range R of the projectile, we multiply the horizontal component of the velocity of projectile with total time taken by the body to hit the ground after leaving the point of projection. Thus; R = vix t As v i x = v i cosθ and t =

2 v i sinθ , so g 2v i sinθ g 2 v 2sinθsinθ R= i g

R = v i cosθ ×

R=

Maximum Range

v i2sin2θ g

………. 11

sinθinθ = sin2θ

The range will be maximum when sin2θ is maximum i.e. Sin2θ = 1

1

Rmax

v i2 = R= g

Also sin90° = 1, so Sin2θ = sin90° Or θ = 45° Thus the projectile must be projected at an angle of 45° to get the maximum range.

Ballistic Flight

A flight in which a projectile is given an initial push and is then allowed to move freely due to inertia and under the action of gravity is called ballistic flight.

Ballistic Missile

An up-powered and un-guided missile is called a ballistic missile.

Ballistic Trajectory

The path followed by the ballistic missile is called ballistic trajectory

Application to Ballistic Missile

As ballistic missile moves in a way that is the result of the superposition of two independent motion. A straight line inertial flight tin the direction of launch and a vertical gravity fall. By law of inertia, an object should sail straight off in the direction thrown, at constant speed equal to its initial speed particularly in empty space. But downward force of gravity will alter straight path into a curved trajectory. For short ranges and flat earth approximation, the trajectory is parabolic but the draggles ballistic trajectory for spherical earth should actually be elliptical. At high speed and for long trajectories the air friction is not negligible and some times the force of air friction is more than gravity. It affects both horizontal as well as vertical motion. Therefore, it is completely unrealistic to neglect the aero-dynamic forces. The shooting of a missile on a selected distant spot is a major element of war force. It undergoes complicated motion due to air friction and wind etc. Consequently the angle of projection can not be found by the geometry of the situation at the moment of launching. The actual flights of missiles are worked out to high degrees of precision and the result was contained in tabular form. The ballistic missiles are useful only for short ranges. For long ranges powered and remote control guided missiles are used. To access these notes please login on www.esnips.com/web/ghulamfarid