introduction

Introduction Mechanics is that branch of science which deals with

the state of rest or motion of bodies under the action of forces. The subject of Mechanics is logically divided into two parts.Statics which concerns the equilibrium of bodies under the action of forces,and Dynamics concerns the motion of Bodies. Dynamics is again subdivided into

a. Kinematics b. Kinetics

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Study of Kinematics concerns with the motion of a body, without referring to the forces causing the motion of that body.

Study of Kinetics concerns with the motion of the body considering the forces causing the motion.

Terms and definitions

space: is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system.

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Time is the measure of succession of events and is basic quantity in dynamics.

Mass is a measure of the inertia of a body, which is its resistance to a change of velocity

Particle. A body of negligible dimensions is called a particle. In the mathematical sense a particle is a body whose dimensions approach to zero so that it may be analyzed as point mass.

Rigid body. A body is considered rigid when the relative movements between its parts are negligible for the purpose at hand.

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Newton’s Laws of Motion

First Law . A particle remains at rest or continues to move in a straight line with a uniform velocity if there is no unbalanced force acting on it.

Second Law. The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force.

Third Law. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.

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Newton’s second law forms the basis for most of the analysis in dynamics. As applied to a particle of mass m, it may be stated as F=ma Where F is the resultant force acting on the particle and a is the resulting acceleration. This equation is a vector equation. Rectilinear Motion: It is the motion of a particle along a straight line.

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A particle moves along a straight line OX as shown in the figure. The moving particle is in position M at any time ‘t’ and it covers a distance ‘s’ from ‘O’. The particle moves to N, through a distance ∆s in a small interval of time ∆t. The velocity v at the instant when the particle is at certain point M, at time ‘t’ is the rate of change of displacement ∆s as the increment of time ∆t approaches to zero as limit is known INSTANTANEOUS VELOCITY and is given by

Lt v = ∆t 0 ∆s/∆t = dS/dt

Rectilinear motion with uniform acceleration:

St

∆s∆t

XM N

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Average Velocity : It is the uniform velocity with which the particle may be considered to be moving in order to cover the total distance s in a total time t.

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(i) When the particle moves with uniform velocity

s = u.t

(ii) When particle moves with variable velocity

𝑉 𝑎𝑣=𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑐𝑜𝑣𝑒𝑟𝑒𝑑 (𝑠 )

𝑇𝑜𝑡𝑎𝑙𝑇𝑖𝑚𝑒 (𝑡 )

(iii) When particle moves with initial velocity u and constant acceleration a, its velocity changes to v, then vav = (u+v)/2

s = (u+v)/2 x t s = distance covered in time ‘t’

(iv) If the distances moved by the particle from start are s1 in t1, s2 in t2, then average velocity may also be found by vavg = (s2 - s1) / (t2-t1)

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If the velocity of a particle is v at M and it changes by ∆v, in a small interval of time ∆t then the acceleration of moving particle, a at the instant at which particle is at M i. e. the instantaneous acceleration is given by

A particle may move in a straight line with constant acceleration or with variable acceleration.

Acceleration : k9

2

2

0 dtsd

dtds

dtd

dtdv

tvLta t

If the velocity of a body changes by equal amounts in equal intervals of time, the body is said to move with uniform acceleration.

Variable Acceleration:

If the velocity of a body changes by unequal amounts in equal intervals of time, the body is said to move with variable acceleration. Note: When the velocity is increasing the acceleration is reckoned as positive, when decreasing as negative (retardation or deceleration) .

Uniform Acceleration :k10

Displacement – Time Variations:

TimeDis

plac

emen

t

Fig. 1

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In fig (1)The graph is parallel to the time axis indicating that the displacement is not changing with time. The slope of the graph is zero. The body has no velocity and is at rest.

Dis

plac

emen

t

Time

∆t∆x

In fig (2)The displacement increases linearly with time. The displacement increases by equal amount in equal intervals of time. The slope of the graph is constant.

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Fig.2

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Dis

plac

emen

t

TimeFig. 3

∆x2∆t2

∆x1∆t1

In fig(3) The displacement is not changing by equal amounts in equal intervals of time. The slope of the graph is different at different times.The velocity of the body is changing with

time. The motion of the body is accelerated.

Velocity – Time Variations:

TimeFig. (a)

velo

city

O

∆v∆t

In fig. (a): The velocity of the body increases linearly with time. The slope of the graph is constant i.e. the velocity changes by equal amounts in equal intervals of time. (acceleration of the body is constant and at t = 0, the velocity is finite. Thus the body moving with a finite initial velocity, and has constant acceleration).

Constant velocity

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Velocity – Time Variations:

Time

Fig. (b)

velo

city

O

∆v∆t

In fig. (b): The body has a finite initial velocity. As time passes, the velocity decreases linearly with time until its final velocity becomes zero i.e it comes to rest. Thus the body at a constant deceleration , since the slope of the graph is negative.

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Uniform acceleration

In fig. (c): The velocity –Time graph is a curve. The slope is, therefore, different at different times. In other words, the velocity is not changing at constant rate. The body does not have a uniform acceleration since acceleration is changing with time.

TimeFig. (c)

velo

city

O

∆v2∆t2

∆v1∆t1

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Variable acceleration

1. Equation of motion (Relation between v,u, a & t)

If we assume a body starts with an initial velocity u and uniform acceleration a. After time t, it attains a velocity v. Therefore the change in velocity in t seconds is (v – u) Change in velocity / sec. = v – u / t = a v = u + at -----(1)

Equations of Motion Under Uniform Acceleration

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2. Equation of Motion: (Relation between s, u, a and t)

Let a body moving with an initial uniform velocity u be accelerated with a uniform acceleration a for time t. If v is the final velocity, the distance s which the body travels in time t is determined as follows.

Now since acceleration is uniform it is obvious that the average velocity = (u + v) /2

Distance traveled = vav x t = (u + v)/2 x t= (u + (u + at))/2 x t (Substituted from 1)

s = ut + ½ at2-----(2)

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3. Equation of motion: (Relation between u, v, a and s) s = average velocity x time

= (u + v)/2 x t

= (u + v)/2 x (v - u)/a for t = (v – u)/a therefore s = (v2 - u2)/2a

v2 = u2 +2as-----------(3)

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Motion under Gravity

It has been observed that bodies falling to the earth (through distances which are small as compared to the radius of the earth) experience entirely unrestricted increase in their velocity by 9.81 m/s for every second during their fall. This acceleration is called the acceleration due to gravity and as conventionally denoted by g.

For downward motion:a = +g v = u + gth = ut + ½ gt2 v2 = u2 + 2gh

For upward motion: a = – g v = u – gth = ut – ½ gt2 v2 = u2 – 2gh

h = Distance moved in vertical direction

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Equations Acceleration 1 2 3 Uniform v = u + at S = ut + ½ at2 v2 = u2 + 2as t t s

Variable v = u+∫ a(t)dt S = ut + 1/2∫ a (t2)dt v2 – u2 = 2 ∫ a (s)ds 0 0 0

Comparison between equations of motion under uniform acceleration and variable acceleration:

The slope ds/dt at any point gives the velocity v at that point.

Graphical Representation:

t1t2 t

dt

ds

v = ds/dt s-t curveS

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The slope dv/dt at any point gives the acceleration a at that point. The shaded area under v – t curve shown above gives the incremental displacement ds during the small interval of time dt.

v

v

dt

dv

dt

t2

v – t curve

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tt1

The shaded area under a – t curve shown above gives the incremental velocity dv during the small interval of time dt.

a

t2t1

dt

dv=adt

t

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PROJECTILES

INTRODUCTION

Assumptions:

1. Mass of the projectile is not considered.

2. Air resistance is neglected.

3. The trajectory of the particle is in the vertical plane.

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A Projectile is a particle moving in space under the action of gravity.

The velocity with which the particle is projected into space has horizontal and vertical components. The combined effect of both components is to move the particle along a parabolic path. The parabolic path traced by the projectile is known as Trajectory of the Projectile.

The horizontal component remains constant ( as air resistance is ignored) while the vertical component of motion is always subjected to acceleration due to gravity.

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DEFINITIONS: Velocity of Projection is the velocity with which a body is

projected into space. Angle of projection is the angle which the initial velocity

vector makes with the horizontal, or the angle at which a projectile is projected with respect to horizontal.

Range is the distance along the reference plane between the point of projection and the point at which it strikes the plane.

Time of Flight is the total time during which the particle remains in motion.

Maximum Height is the maximum vertical distance covered by the projectile from the point of projection.

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MOTION OF A PROJECTILE

u

R

trajectory

x

y

O

P(x,y)

M

Consider a particle thrown upwards from a point O, with an initial velocity u, at an angle with the horizontal as shown in the figure above. After attaining maximum height h, it descends and finally hits the reference plane.

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h

Q

Equation of the Trajectory

From the figure above

x = ux . t = u (Cos ) t

t = x / (u Cos ) --------------(1)

y = uy. t – ½ gt2

=u (Sin )t – ½ gt2------------------(2)

Sub (1) in (2) we get

y = x tan - gx2/ (2u2 Cos2 )

This is an equation for a parabola. Hence the path of the projectile is a parabola.

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The horizontal distance covered by the projectile is known as Range of Projectile denoted by R.Resolving the initial velocity u into horizontal and vertical components ux = u Cos uy = uSin

(constant)

Time of Flight:We know that vy = uy + atAt Q vy = 0 0 = uy - gt

tm = uy / g = u Sin / g Time of flight T = 2tm = 2 (u Sin /g)

Where tm is the time taken in seconds to reach maximum height.

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Maximum height attained (h):

vy2 - uy

2 =- 2gh (upward motion)

vy = 0 h = uy2/2g ;

h = u2(Sin2) / 2g

Range:

R= ux x T (time of flight)

= 2u2(SinCos) / g

R = u2Sin2/g (Sin2 = 2SinCos)

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From the above equation it is clear that range will be maximum if Sin2 = 1

i.e. 2 = 90 or = 45 Rmax = u2/g

Projectile will cover a maximum range when it is directed at an angle of 45°.Two angles of Projections for a given range:We know thatRange R = u2Sin2/g Sin = Sin ( - )u2Sin ( - 2)/g = u2Sin (2 1)/g (say) where 21= ( - 2)

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Thus for same range R = u2Sin2/g = u2Sin ( - 2)/g = u2Sin21/g

Which shows that the horizontal range remains the same when is replaced by 1.

Or 1= - 2 =( / 2 )-

2

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Projection on an Inclined Plane

u

R

B

O

R(Sin)

Q

R(cos)

R= Range along incline; α=angle of projection;

β =angle of inclined plane

Sx=u x+ ½ axt2

P

P1

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Projection on an Inclined Plane

R(cosβ)=u(cosα)(t)+(0) ; ax =0

t= Rcosβ / ucosα…..(1)

Sy=uy + ½ ayt2

R sinβ=u sinα (t)- ½ gt2

Substituting eqn.(1) and simplifying we get

The Range along the inclined plane

R={2u2cos2 α [sin(α- β)] } / gcos2 β ..(2)

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Projection on an Inclined Plane:

To get maximum range on the incline,

Differentiating R w.r.t α and equating it to zero

we get α= π/4 + β/4

Substituting this value of α in eqn.(2)

We get maximum Range

Rmax= u2/g(1+sin β)

To find the time of flight: using the relation

v=u + at

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0=u x sin(α- β)-gx(cos β) x t

t= {usin(α- β)}/ g(cos β)

a=(g cos β) is the acceleration due to gravity along inclined plane

t=time taken by the projectile to reach Q where QP is perpendicular distance to the incline plane

Time of Flight T= 2xt

T={2 u sin(α- β)} / g(cos β)

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Maximum Height attained h, (PQ)

using the relation

v2-u2=2as

0-u2 sin2(α- β)=-2 gcos β x h

h={ u2 sin2(α- β)} / 2gcos β

Vertical height P1Q=h/ cos β

={u2 sin2(α- β)} / 2gcos2 β

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The distance between two stations is 3km. A Locomotive starting from one station gives the train an acceleration of 1/3 m/s2 until the speed reaches 72kmph.Then the speed is maintained until the brakes are applied giving a retardation of 1m/ s2 and the train is brought to rest at the next station. Find the time taken to perform the journey and distances covered during acceleration,constant speed and retardation

Problem: 1Linear motionk38

solution:

v

t

S1S2

S3

a. Using V-T curve

B D

a1=1/3m/s2; v=72kmph=20m/s

Therefore t1= v/a1 =60s (v/t=a)

Similarly t3=20s

t1 t2 t3

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solution:

S1=1/2 x v x t1 = 600m

s3=1/2 x v x t3 = 200m

S2 = 3000-200-600=2200m

t2= 2200/20=110s

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s1 ,t1 s2 ,t2s3 ,t3

A B C D

s1+ s2 + s3=3000m

From A to B,

vB=20=0+1/3t1 (v = u+at)

Alternate methodk41

t1 =60 s

From C to D,

0 = 20- 1x t3

t3 = 20s

s3 = 20 x 20-0.5x 1 x 202

s3 = 300m

Therefore s2 =3000-600-200

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s2= 2200m; t2= 2200/20 = 110s

Problem 2

A burgler’s car had a start with an acceleration of 2m/s2. A police vigilance party came after 5 seconds and continued to chase the burgler’s car with a uniform velocity of 20m/s. Find the time taken ,in which the police will overtake the car.

Solution: Let the police party overtake the burgler’s car in t seconds, after the instant of reaching the spot.

Distance travelled by the burglar’s car in t1 seconds is s1

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u=0, a = 2m/ s2

Time, t1 = (5 + t) secs.

s = ut + 0.5at2

s1= 0 +0.5 x 2 x (5+t) 2

= (5+t) 2

Distance travelled by the police party,

s2= v x t

( police moves with uniform velocity )

Solution: k44

a=2m/s2

v=20m/s

Bugler’s car

police

At the time of overtaking,

the distances covered should be equal

s1= s2

(5+t)2 = 20t

Solving the above quadratic equation

k45

we get t=5s

Two trains A and B leave the same station on parallel lines.A starts with uniform acceleration of 0.15m/s2 and attains a speed of 24kmph when the steam is reduced to keep speed constant. B leaves 40 seconds later with uniform acceleration of 0.30m/s2 to attain a maximum speed of 48km/hr. When will B overtake A ?

Problem 3k46

Motion of train A:

Uniform Acceleration, a1= 0.15 m/s2

Initial Velocity , u1= 0

Final Velocity , v1 = 24kmph

= 20/3 m/s

Let t1 be the time taken to attain this velocity(in seconds)

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Using the relation:

v = u + at

20/3 = 0 + 0.15 x t1

Therefore, t1 = 20/3 x 0.15 = 44.4 s

Also, distance travelled during this interval,

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s1= u1t1 + ½ a1t12

= 0 + ½ x 0.15 x 44.42 = 148 m.

Motion of Train B:

u2 = 0

a2 = 0.3 m/s2

v2 = 48 kmph= 40/3 m/s

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Let t2 be the time taken to travel this distance,say s2,

Using the relation:

v = u + at

40/3 = 0 + 0.3 x t2

Therefore, t2 = 40/3x0.3 = 44.4 s.

And s2= u2t2 + ½ a2t22

=0+ ½ x0.3x(44.4)2=296m

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Let the train B overtake the train A when they have covered a distance s from the start. And let the train B take t seconds to cover the distance.

Thus, time taken by train A = (t+40)s

Total distance moved by train A,

s = 148+distance covered with constant speed.

= 148 + [(t+40)-t1]x 20/3

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= 148 + [t+40-44.4] x20/3.

= 148 + (t – 4.4) x20/3. …….(i)

[{t+40) – t2} is the time during which train A moves

with constant speed]

Similarly,total distance travelled by train B,

s = 296+ distance covered with constant speed.

=296 + ( t-44.4)x 40/3 ………(ii)

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equating (i) and (ii)

148 + (t-4.4)x20/3 = 296 + (t-44.4) x40/3

148 + 20/3t –88/3 = 296 +40/3t –1776/3

(40/3-20/3)=148-296+1776/3-88/3

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t = 62.26s

Hence, Train B overtakes train A after 62.26s after its start.

where t = v / a = 6.66 / 0.15

= 44.4ses

Using v -t curve:( for train A)

sA=0.5 x44.4 x6.67+6.67 x(t+40-44.4)…. 1

a=0.15m/s2

v=6.67m/s(24kmph)

v

t

k54

where t2=v/a=13.3/0.30

=44.4ses

Using V-T curve: ( for train B)

sB=0.5 x44.4 x13.33+13.33 x(t-t2)…. 2

a=0.30m/s2

V =13.33m/s(48kmph)

v

tt240s

At the time of overtaking sA=sB ;

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t=62.2s

Vertical motion

Problem 4

A stone dropped into a well is heard to strike the water after 4 seconds. Find the depth of the well, if the velocity of sound is 350m/s.

u = 0 ; h = depth of well

t=Time taken by the stone to strike the water level of the well.

Then h = ut + 0.5gt2

= 4.905t2 ……… (1)

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Also time taken by the sound to reach the top= depth of well/Velocity of sound

= h/350 = 4.905t2 / 350

Total time taken = time taken by the stone to reach the water level of the well + time taken by sound to reach the top of the well = 4s (given)

. . . t + 4.905t2 / 350 = 4

4.905t2 + 350t -1400=0

Solving for t,

k57

we get t= 3.8s and h=70.8m (from eq.1)

A Cage descends a mine shaft with an acceleration of 1m/s2. After the cage has travelled 30m, a stone is dropped from the top of the shaft.

Determine:

(i) the time taken by the stone to hit the cage, and

(ii) distance travelled by the cage before impact.

Problem 4k58

acceleration of cage,

a= 1m/s2

Distance travelled by the shaft before dropping of the stone=30m

(a) Time taken by the stone to hit the cage = ?

Considering the motion of the stone.

Initial velocity, u = 0

Solution: k59

Let t= time taken by the stone to hit the cage,

and

h1= vertical distance travelled by the stone

before the impact.

Using the relation,

h = ut + ½ gt2

h1 = 0 + ½ x 9.81 t2 = 4.905 t2 ………(i)

Now let us consider the motion of the cage for 30m.

k60

Initial Velocity u = 0

acceleration a = 1.0 m/s2

Let t1 = time Taken by the shaft to travel 30 m.

Using the relation,

s = ut +1/2 at2

30 = 0+1/2 x 1 x (t1)2

t1=7.75 s.

It means that cage has travelled for 7.75s, before the stone was dropped.

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Therefore the total time taken by the cage before the impact = (7.75 + t).

Again using the relation:

s = ut + ½ at2

s1 = 0 + ½ x 1 x (7.75 + t )2……..(ii) In order that stone may hit the cage the two distances must be equal,that is, equating (i) and(ii) and solving for t we get

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t = 2.75 s.

(b) Distance travelled by the cage before impact=?

let s2= distance travelled by the cage before impact.

We know total time taken by the cage before impact.

= 7.75 +2.75 =10.5 s.

Now using the relation,

s2 = ut + ½ at2= 0 + ½ x 1 x (10.5)2

k63

Hence, distance travelled by the cage before impact = 55.12 m.

If a body moves in a straight line and its distance s is in meters, from a given point in the line, after t seconds ,is given by the equation

s=20t + 3t2 –2t3

Calculate:

a. The velocity and acceleration at the start

b. The time when the particle reaches its

maximum velocity

c. The maximum velocity of the body

Problem 5Variable acceleration k64

The equation of motion is: s = 20t + 3t2 – 2t3 …..1

ds/dt = v = 20 + 6t - 6t2 …...2

d2s/dt2 = dv/dt = a = 6 -12t …...3

at start t=0, v = 20 + 0 – 0 = 20m/s

a = 6 –12 x 0 = 6m/s2

When the particle reaches maximum velocity

a=0 ; i.e 6 -12t = 0 ;

k65

t = 0.5secs

When t = 0.5 secs: vmax= 21.5m / sec(sub. t in eqn 2)

A body moves in a straight line has the equation of

motion given by s = 2t3 - 4t + 10

Determine:

i) The time required for the body to reach a velocity of 68m/s starting from rest

ii) The acceleration of the body when velocity is equal to 32m/s

iii) The net displacement of the body between the time interval of t = 0 to t = 4s

Problem 6

k66

s = 2t3 - 4t + 10…….1

Then ds/dt = v = 6t2 - 4……..2

dv/dt=a=12t……3

(i) Time required for the body to reach a velocity v= 68m/s,

Solution: k67

t = 3.46m/s (ii) To find out the acceleration we should know t when

v = 32m/s.

From eqn. 2, v = 6t2 – 4

32 = 6t2 - 4

t = 2.45 s

Substituting this value of t in eq.3, we get

a = 12t = 12 x 2.45

k68

(iii) Net displacement of the body between time t = 0 and t = 4s. , Snet = ?

Substituting values for t in eq. (1), we get

S(t=0) = s1 = 2x0- 4 x 0 + 10 = 10 m.

S = 10 m .

a= 29.4 m/s2

It means that the body is 10 m ahead of starting point. Hence 10m is to be deducted from the final displacement when t = 4s.

S(t=4) = s2 = 2 x (4)3 – 4 x 4 + 10 = 122 m

Snet = (s2 – s1) = 122 – 10 =112 m (ans.)

k69

Problem 7

A train travels along a straight track such that the distance travelled is directly proportional to the cube of the time of interval. If it travels 216m during the first minute,find the displacement,velocity and acceleration just at the start and 10 seconds after starting.

k70

s α t3

s = Ct3……………(1) s = distance travelled, t = time taken, C= constant = 216m =60secs

Substituting the values in equation (1) 216=Cx(60)3 C=0.001S= 0.001t3 …………….(i)

ds/dt =v=0.003t2…………….(ii) dv/dt =a=0.006t…………….(iii) At the start ie at t=0; displacement s= 0 ; v=0; a=0[substituting t=0 in (i) ,(ii) ,(iii)]10 seconds after starting(substitute t=10s)

Solution: k71

S=1m, v=0.3m/s, a= 0.06m/s 2

A rifle bullet is fired at and hits a target 100 m distant. If the bullet rises to a maximum height of 3 cm above the horizontal line between the muzzle of the rifle and the point of impact on the target which is as same level as the muzzle of the rifle, find the muzzle velocity.

Problem: 8PROJECTILESk72

100 m

3cmX

Y

u

O

Solution:k73

Let u = velocity of projection of the bullet α = angle of projection ( with the horizontal )

Now maximum height, h= u2 sin2 α = 3 cm =0.03 m 2g

u2 sin 2α = 0.03 ………...(i) 2g

Also horizontal range, r = u2 sin 2α = 100 g

k74

u2 sin 2α = 100 ………..(ii) g Dividing (i) by (ii), we get u2 sin 2α = 0.032u2 sin 2α 100

Sin 2 α = 0.0003 2x2 sin α cos α

tan α = 0.0012

k75

Since angle α is very small, sin α = tan α = 0.0012Now, from equation (i) u2 x (0.0012)2 = 0.03 2 x 9.81 u2 = 0.03 x 2 x 9.81 = 408750 (0.0012)2

k76

u= 639.3 m/s.

A missile projected from a launching point had reached a vertical distance of 20 km and horizontal distance of 10km from the launching point, when it was found that the fuel of the rocket was exhausted. The missile had then acquired a velocity of 1600 m/s. at an angle of 35° with the horizontal. Assuming that the rest of the flight is under the influence of gravity and neglecting air resistance and curvature of the earth, calculate:

i) Time of flight after the fuel had completely burnt, and

ii) Total horizontal range from the launching point.

Problem 8 k77

Solution:

10km

20km

35°h

Qu = 1600m/s

Trajectory

O M

Velocity of the rocket when the fuel is exhausted.

u = 1600m/s

P

k78

i) Time of flight after the fuel had completely burnt:

Let O be the launching point, from where the missile is projected. Similarly, let P be the point where the fuel was found to be exhausted.

Let t1= time taken by the missile to reach Q i.e. maximum height from P, and

t2= time taken by the missile to reach the ground M from Q

k79

Considering motion of missile from P to Q, t1= (u sin ) / g = 1600 sin 35°/9.81 = 93.6 s.(or using vy= uy-gt)And maximum height to which the missile will rise from P,h = (u2 sin2 )/ 2g (or using sy=uyx t-o.5 x g x t2) = (16002 x sin2 35°) / (2 x 9.81)

k80

Now considering vertical motion of the missile from Q to R.

We know that total vertical distance

= h + 20000= 42926 + 20000 = 62926 m

= 42926m

And initial velocity, u = 0.Using the equation:

62926 = ut + ½ gt22= 0 + ½ x 9.81 x t2

2

t2 = 113.3 s.

Total time , T = t1 + t2

= 93.6 + 113.3 = 206.9 s.

k81

ii) Total horizontal range from the launching point,

R = 10000 + 1600 cos 35° x 206.9

T= 3min. 27 s.

Range= 281172 m = 281.172 km

Problem 9

A bullet is fired upwards at an angle of 30° to the horizontal from a point P on a hill and it strikes a target which is 80m lower than P. The initial velocity of the bullet is 100 m/s.Calculate i) The maximum height to which the bullet will rise above the point of projection.ii) The actual velocity with which it will strike the target.iii) The total time required for the flight of the bullet.iv) Horizontal distance between the point of firing and the target.

k82

Solution:

Horizontal distance(s)

=30°

trajectory

h

Quy

P M

u = 100m/s

80m targethill

v

Vx

vy

ux

L

y

x

k83

Given that:Initial velocity, u = 100 m/sAngle of projection, = 30°Depth to the target = 80 mHorizontal component of initial velocity uux = u cos = 100 cos 30° = 86.67 m/sVertical component:uy = u sin = 100 sin 30° = 50 m/s

i) Maximum height to which the bullet will rise, h = ?Let the maximum height (above horizontal through P) to which bullet will rise be h as shown.

k84

Considering motion along Y axis:Using the relation:

h = (u2 sin2 ) / 2g = (1002 x sin2 30°) / (2 x 9.81) = 127.42m.ii) Actual velocity with which the bullet hits the target = ?

In order to determine vy using the relation v2

y - u2y = 2g x 80

v2y – 502 =- 2 x 9.81 x 80

vy = 2500 + 2 x 9.81 x 80

k85

= 63.8 m/s

Resultant velocity at L

v = v2y + v2

x

But vx = ux = 86.67 m/s.

v = 63.82 + 86.672 =

k86

iii) Total time required for the flight of the bullet, t = ?

using vy = uy-gt

-63.8 =50 -9.81 x t

107.6 m/s

t =113.8/9.81 =11.6 sec

iv) Horizontal distance between point of firing and the target, s = ? s = ux x t

= 86.67 x 11.6

k87

= 1005.4 m

Problem 10

A particle is projected from a point, on an inclined plane, with a velocity of 30 m/s. The angle of projection and angle of plane are 55° and 20° to the horizontal respectively. Show that the range up the plane is the maximum for the given plane. Find the range and time of flight of the particle.

k88

Velocity of projectile, u = 30 m/s.

Angle of projection, = 55°

Angle of plane, = 20°

i) Condition for maximum range = ?

We know that for maximum range, the angle of projection,

= /4 + /2 = 180/4 + 20/2 = 55°

Since the given angle of projection is 55°, therefore range up the plane is the maximum for the plane (Ans.)

Solution: k89

ii) Range of projectile, R = ?

Using the relation:

R = (2u2 sin ( - ) cos )/ (g cos2 )

= (2 x 302 sin (55° - 20°) cos 55° ) / 9.81 cos2 20°

= (2 x 900 x 0.573) / 9.81 x 0.883

k90

= 68.22 m

iii) Time of flight, t = ?

Using the relation:

t = (2u sin ( - ))/ g cos

= (2 x 30 sin (55° - 20°)) / 9.81 cos 20°

= (2 x 30 x 0.573) / 9.81 x 0.9396

k91

= 3.73 s

1. On turning a corner, a motorist rushing at 15m/s, finds a child on the road 40m ahead. He instantly stops the engine and applies brakes,so as to stop the car within 5m of the child, calculate: (a) retardation (b) time required to stop the car

Ans: (a) -3.21m/s2, (b) 4.67s

2. A stone is dropped from the top of a tower 100m high. Another stone is projected upward at the same time from the foot of the tower, and meets the first stone at a height of 40m. Find the velocity,with which the second stone is projected upwards.

Ans: u=28.6 m/s

k92Practice Problems:


3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph.

Ans: S=293m

4. A projectile is fired from the edge of a 150m high cliff with an initial velocity of 180m/s at an angle of elevation of 30o with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground,and (b) the greatest elevation above the ground reached by the projectile.

(a) 3125m (b) 563m

k94

Practice Problems:

3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph.

Ans: S=293m

4. A projectile is fired from the edge of a 150m high cliff with an initial velocity of 180m/s at an angle of elevation of 30o with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground,and (b) the greatest elevation above the ground reached by the projectile.

(a) 3125m (b) 563m

k95

Practice Problems:

5. An aero plane is flying at a height of 300m,with a velocity of 360kmph. A Shell is fired from the ground exactly when the aeroplane is above the gun.what should be the minimum initial velocity of the shell and the angle of inclination in order to hit the aeroplane ?

Ans:( 126m/s, 37.47o)

6. A projectile is fired from a point ‘o’ at a velocity of 125m/s has to strike a point located on the top of a tower of 200m high. The horizontal distance betweem the point ‘o’ and the tower is 1000m. Neglect the air resistance and take g= 9.8m/s2


Contd:

Calculate:

a) The angel, to the horizontal,at which the projectile must be fired in order to strike the point on tower in minimum time.

b) the time taken for flight

c) The maximum height above ‘o’ reached by the projectile.

Ans: (a) 68.550 or 32.990 ,( b) 9.54 s, (c) 233.8m