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CHAPTER 2

KINEMATICS IN ONE DIMENSION

Study Objectives:

At the end of the class, students should be able to

a) define distance, displacement, speed, velocity and acceleration

b) state the difference between vector and scalar quantities

c) solve problems using equations of motion with constant acceleration

d) sketch and interpret graphs of displacement-time, velocity-time and acceleration-

time for motion of a body

e) solve problems using equations of motion with constant acceleration to analyze free

fall.

Mechanics: the study of how objects move and the forces that causes motion.

Dynamics: the branch of physics that studies force and the causes of various types of motion.

Kinematics: the branch of physics that describes motion of objects without considering the

effects that produce the motion. The motion of objects can be explained by using words,

diagrams, numbers, graphs, and equations. The goal of any study of kinematics is to develop

sophisticated mental models which serve to describe (and ultimately, explain) the motion of real-

world objects.

2.1 DISPLACEMENT

Motion is related to change of position.

Distance and displacement are two quantities which may seem to mean the same thing, yet they

have distinctly different meanings and definitions.

Distance

– The actual path length between two points/length measured along the path line.

- Distance is a scalar quantity (magnitude with no direction)

Displacement

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– The displacement is a vector that points from an object’s initial position to its final

position and has a magnitude that equals the shortest distance between the two positions.

– The direct straight line pointing from the initial point to the final point / the change in position

of the object.

- Displacement is a vector quantity (has both magnitude & direction)

- The displacement: 0xxx

- + sign: direction to the right/east

- - sign: direction to the left/west

- SI Unit : meter (m)

Fig 2.1 The arrow represents the displacement x2 x1 = x .

Figure 2.1 The displacement x is a vector that points from the initial position x0 to the final

position x.

EXAMPLE 2.1:

Distance = 3.0 m + 4.0 m = 7.0 m

Displacement = 5.0 m

C

5.0 m

A 3.0 m

B

4.0 m

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EXAMPLE 2.2

A boy runs 30 m east, 40 m north, 50 m west. What is his net displacement?

SOLUTION :

mmmmd 45)3050()40( 22

01 2740

3050tan

m

mm west of north

AVERAGE SPEED – indicates how fast an object is moving.

The average speed of an object is the total distance traveled by the object per unit time.

Average speed = timeElapsed

traveldistance total

Speed is a scalar quantity.

SI unit : m/s

AVERAGE VELOCITY - indicates how fast an object is moving and the direction of its

motion.

Defined as displacement of the object divided by the time interval during which the

displacement occurred.

Average velocity = takentime

ntdisplaceme of change

t

x

tt

xx

0

0

Average velocity is a vector quantity and its direction follows that of the change in displacement.

SI unit : m/s

50 m

30 m

40 m

d

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INSTANTANEOUSLY VELOCITY

- is the velocity (magnitude & direction) of an object at a particular instant of time ( t is closed

to zero)

t

x

t

0lim

SI unit : m / s

Instantaneous Speed - speed at any given instant in time.

Average Speed - average of all instantaneous speeds; found simply by a distance/time ratio.

EXAMPLE 2.3 :

A plot of position versus time is in Fig. 1 for an object in linear motion.

gure 1

(a) What are the average velocities for the segments AB,BC,CD,DE,EF,FG, and BG?

(b) State whether the motion is uniform or non uniform in each case.

(c) What is the instantaneous velocity at point D?

SOLUTION:

(a) if

if

tt

xx

t

x

000.1

0.10.1

s

mmAB

smss

mmBC /0.3

0.10.3

0.10.7

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smss

mmCD /3.1

0.35.4

0.70.9

smss

mmDE /3.1

5.40.6

0.90.7

smss

mmEF /7.1

0.60.9

0.70.2

00.90.11

0.20.2

ss

mmFG

smss

mmBG /10.0

0.10.11

0.10.2

b) The motion of BC, CD, and DE are not uniform since they are not straight lines.

c) The object changes its direction of motion at point D. So it has to stop momentarily and

0 .

ACCELERATION

When the velocity of an object changes in magnitude, or in direction, or in both magnitude and

direction, the object is said to accelerate.

Acceleration is the rate of change of velocity.

Average acceleration the change in velocity divided by the time interval to make the

change.

timeelapsed

yin velocit changeonaccelerati average

0

0

ttta

Acceleration is a vector quantity and its direction follows the direction of the change in velocity.

The direction of the acceleration vector depends on two factors:

whether the object is speeding up or slowing down

whether the object is moving in the positive (+) or negative (–) direction

a = 0 (zero acc ) ---velocity is a constant

a = + (positive acc) --- acceleration (speed increase)

a = - (negative acc) --- deceleration (speed decrease)

Instantaneous acceleration is the acceleration of an object at a particular instant of time.

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SI unit : m/s2

The direction of the instantaneous acceleration follows the direction of the instantaneous velocity

change.

2.2 Equations Of Kinematics For Constant Acceleration (Uniform acceleration)

A body can have many ways of motion: in one dimensional path, curve, circular, parabolic and

many others. This chapter will only discuss the motion on one straight line with a constant

acceleration.

Kinematics variables: 1) x = displacement

2) a = acceleration

3) = final velocity at time t

4) 0 = initial velocity at time t0=0 s

5) t = time elapsed since t0=0 s

Kinematics Equations:

The equations of kinematics apply when an object moves with a constant acceleration along a

straight line. These equations relate the displacement x-x0, the acceleration a, the final velocity

, the initial velocity 0 , and the elapsed time t-t0.

Assuming that x0=0 m at t0=0 s, the equations of kinematics are:

1) at 0

2) 2

0

3) tx )(2

10

4) 2

02

1attx

5) ax22

0

2

The process to determine unknown information about an object's motion involves the use of a

problem-solving strategy which includes the following steps:

1. Construct an informative diagram of the physical situation.

2. Identify and list the given information in variable form.

3. Identify and list the unknown information in variable form.

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4. Identify and list the equation which will be used to determine the unknown information

from the known variables.

5. Substitute known values into the equation and use appropriate algebraic steps to solve for

the unknown.

6. Check your answer to ensure that it is reasonable and mathematically correct.

EXAMPLE 2.4

A car accelerates from 13m/s to 25m/s in 6.0s.What was it acceleration? How far did it travel in

this time? Assume constant acceleration.

Solution:

By definition, the acceleration is 20

25m s 13m s2.0m s

6.0 s

v va

t

.

The distance of travel can be found from Eq. 2-11b.

22 21 1

0 0 2 213m s 6.0 s 2.0m s 6.0 s 114 mx x v t at

EXAMPLE 2.5 A car slows down from 23m/s to rest in a distance of 85m. What was it acceleration, assumed constant?

Solutions:

22 2

2 2 20

0 0

0

0 23m s2 3.1m s

2 2 85 m

v vv v a x x a

x x

.

EXAMPLE 2.6 A world class sprinter can burst out of the blocks to essentially top speed(of about 11.5m/s) in the first

15.0m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach

that speed?

Solutions:

The sprinter starts from rest.

22 2

2 2 2 20

0 0

0

11.5m s 02 4.408m s 4.41m s

2 2 15.0 m

v vv v a x x a

x x

.

The elapsed time is found by solving Eq. 2-11a for time.

0

0 2

11.5m s 0 2.61 s

4.408m s

v vv v at t

a

2.3 Motion Diagrams / Graphical Analysis

- Graphical techniques are often helpful in understanding motion and its related quantities

especially velocity and acceleration.

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bmxy : ta 0

Graph : position versus time (x-t) --- velocity (slope)

Velocity versus time (v-t) --- acceleration (slope)

(A) Displacement graphs

Case 1: Positive velocity, constant velocity:

To begin, consider a car moving with a constant, rightward (+) velocity of 10 m/s.

Note that a motion with constant, positive velocity results in a line of constant and positive slope

when plotted as a position-time graph.

Figure (a) : Positive Velocity Constant Velocity

Conclusions:

1) Instantaneous velocity is the same at any time.

2) The gradient is positive which means positive velocity. Horizontal line, velocity is zero.

Case 2: Positive velocity, changing velocity (acceleration)

Now consider a car moving with a changing, rightward (+) velocity – that is, a car that is moving

rightward and speeding up or accelerating.

If the position-time data for such a car were graphed, the resulting graph would look like the

graph as shown above. Note that a motion with changing, positive velocity results in a line of

changing and positive slope when plotted as a position-time graph.

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Figure (b) : Positive Velocity Changing Velocity (acceleration)

Conclusions:

The velocity is increasing because the gradient is increasing

Other types of graph :

Leftward (–) Velocity; Slow to Fast Leftward (–) Velocity; Fast to Slow

B. Velocity Graphs

Figure (a) – the object moving with a positive velocity

Figure (b) – the object moving with a negative velocity

The method used to find the area under a line on a velocity-time graph depends on whether the

section bounded by the line and the axes is a rectangle, a triangle or a trapezoid.

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The shaded area is representative of the distance traveled by the object

during the time interval from 0 seconds to 6 seconds. This

representation of the distance traveled takes on the shape of a

rectangle whose area can be calculated using the equation:

Rectangle : Area = b x h

The shaded area is representative of the distance traveled by the object

during the time interval from 0 seconds to 4 seconds. This

representation of the distance traveled takes on the shape of a triangle

whose area can be calculated using the equation:

Triangle : Area =2

1 x b x h

The shaded area is representative of the distance traveled by the object

during the time interval from 2 seconds to 5 seconds. This

representation of the distance traveled takes on the shape of a

trapezoid whose area can be calculated using the equation:

Trapezoid : Area = 2

1x b x (h1 + h2)

EXAMPLE 2.7

A bicyclist maintains a constant velocity on the outgoing leg of a trip, zero velocity while

stopped, and another constant velocity on the way back as shown below. Obtain the velocities for

each segment of the trip.

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EXERCISE 2.8:

REASONING AND SOLUTION The average acceleration for each segment is the slope of that segment.

2.4 FREELY FALLING OBJECTS

A freely falling body is an object moving freely under the

influence of gravity only, regardless of its initial motion (in the

absence of air resistance).

All objects at the same location above the earth have the same

acceleration due to gravity. The acceleration due to gravity is

directed toward the center of the earth and has the magnitude of

approximately g = 9.80 m/s2 near the earth surface.

Near the surface of the Earth, g = 9.80 m/s2

Near the surface of the Moon, g = 1.70 m/s2

The Big Misconception

Figure 2.16 (pg 40) – (a) In the presence of air resistance, the acceleration of the rock is

greater than that of the paper.

(b) In the absence of air resistance, both the rock and the paper have

the same acceleration.

aA

= 40 m/s - 0 m/s

21 s- 0 s =

aB

= 40 m/s - 40 m/s

48 s - 21 s =

aC

= 80 m/s - 40 m/s

60 s - 48 s =

0 m/s2

3.3 m/s2

1.9 m/s2

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How Fast? and How Far? – Equations of kinematics

Since free fall is in vertical direction, we often choose :

the upward direction as the +y axis,

replace the : x’s y’s and

a’s –g’s (a=-g=-9.8)

Since the acceleration is constant in free-fall, the equations of kinematics can be used.

1. ty

2. 2

0

3. gt 0

4. 2

02

1gtty

5. gy22

0

2

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The application of these four equations to the motion of an object in free fall can be aided by a

proper understanding of the conceptual characteristics of free fall motion. These concepts are as

follows:

1. An object in free fall experiences an acceleration of –9.8 m/s2. (The negative (–)

sign indicates a downward acceleration). Whether explicitly stated or not, in the

kinematics equations the acceleration for any freely falling object is ALWAYS –

9.8 m/s2.

2. If an object is dropped (as opposed to being thrown) from an elevated height to

the ground below, the initial velocity of the object is 0 m/s.

3. If an object is projected upwards in a vertical direction, it will slow down as it

rises upward. The instant at which it reaches the peak of its trajectory, its velocity

is 0 m/s. This value can be used as one of the motion parameters in the kinematics

equations; for example, the final velocity (vf) after traveling to the peak would be

assigned a value of 0 m/s.

4. If an object is projected upwards in a vertical direction, then the velocity at which

it is projected is equal in magnitude and opposite in sign to the velocity it has

when it returns to the same height. That is, a ball projected vertically with an

upward velocity of +30 m/s will have a downward velocity of –30 m/s when it

returns to that same height.

These four principles and the four kinematics equations can be combined to solve hproblems

involving the motion of free falling objects.

Analysis of the slope:

Position vs. Time Graphs

The position vs. time graph for a free-falling object is shown below.

A curved line on a position vs. time graph signifies an accelerated motion. A closer look at the

position-time graph reveals that the object starts with a small velocity (slow) and finishes with a

large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object,

the initial small slope indicates a small initial velocity and the final large slope indicates a large

final velocity.The negative slope of the line indicates a negative (i.e., downward) velocity.

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Velocity vs. Time Graphs

The velocity vs. time graph for a free-falling object is shown below.

A diagonal line on a velocity vs. time graph signifies an accelerated motion. A closer look at the

velocity-time graph reveals that the object starts with a zero velocity (starts from rest) and

finishes with a large, negative velocity; that is, the object is moving in the negative direction and

speeding up. An object which is moving in the negative direction and speeding up is said to have

a negative acceleration. Since the slope of any velocity vs. time graph is the acceleration of the

object, the constant, negative slope indicates a constant, negative acceleration. This analysis of

the slope on the graph is consistent with the motion of a free-falling object – an object moving

with a constant acceleration of 10 m/s/s in the downward direction.

EXAMPLE 2.9:

A ball is thrown upward with an initial velocity of 10.0 m/s from the top of a 50.0 m tall

building.

(a) With what velocity will the ball strike the ground?

(b) How long does it take the ball to strike the ground?

SOLUTION:

Given : y = -50.0 m (displacement) v0 = +10.0 m/s

Find : (a) t and (b) v

The y in the kinematics equations stands for displacement from the launch point, not distance.

When the ball strikes the ground, it will displace

y = -50.0 m

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-50.0 m, or 50 m below the launch point.

(a) 223

222

0

2

/1008.1

)0.50)(/80.9(2)/0.10(2

smx

msmsmgy

So smsmx /9.32/1008.1 223

The +ve answer is discarded since the ball is falling when it lands (moving downward).

EXAMPLE 2.10

A stone is thrown vertically upward with a speed of 12.0m/s from the edge of a cliff 70.0 m high.

a) How much later does it reach the bottom of the cliff?

b) What is its speed just before hitting?

c) What total distance did it travel?

Solutions:

Choose downward to be the positive direction, and 0

0y to be at the top of the cliff. The initial

velocity is 0

12.0m sv , the acceleration is 2

9.80m sa , and the final location is 70.0 my .

(a) Using Eq. 2-11b and substituting y for x, we have

2 2 21

0 0 2 4.9m s 12.0m s 70 m 0 2.749 s , 5.198 sy y v t at t t t .

The positive answer is the physical answer: 5.20 st .

(b) Using Eq. 2-11a, we have 2

012.0m s 9.80m s 5.198 s 38.9m sv v at .

(c) The total distance traveled will be the distance up plus the distance down. The distance down

will be 70 m more than the distance up. To find the distance up, use the fact that the speed at

the top of the path will be 0. Then using Eq. 2-11c:

22 2

2 2 0

0 0 0 2

0 12.0 m s2 0 7.35 m

2 2 9.80m s

v vv v a y y y y

a

.

Thus the distance up is 7.35 m, the distance down is 77.35 m, and the total distance traveled is

84.7 m .

KINEMATICS IN 2-D

2.5 DISPLACEMENT, VELOCITY AND ACCELERATION

In two dimensions, it is necessary to use vector notation to describe physical quantities with both

magnitude and direction. In this chapter, we define displacement, velocity and acceleration as

vectors in two dimensions

i) Displacement, r

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Displacement is difference in the position vectors from the initial point at time to to

final point at time t.

Consider the motion of the car below in figure 1. Two of the positions are identified with

the vectors r and ro. These are drawn from an origin of our choice

The displacement of the car, r, is drawn from the initial position at to and final position

at t. From the drawing it is easy to see that r = ro + r and the displacement is given by

r = r - ro

ii) Average Velocity, v

Average velocity of an object is the ratio of the displacement to the time interval for this

displacement.

Now the average velocity of the car between the two positions is just the displacement

divided by the time interval:

v = o

o

r r

t t

=

r

t

iii) Instantaneous velocity, v

Figure 1

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Instantaneous velocity is the velocity at a particular point in time.

Δr is measured during a very short increment of time and divided by that increment

of time.

Or the instantaneous velocity is merely the average velocity where the elapsed time is

very short.

lim Δrv =

Δt 0 Δt

On the curved part of the track (figure 2), the instantaneous velocity vector is tangent to

the curve, since that is the direction of the car’s motion at any instant in time.

In other words, the instantaneous velocity is parallel to the tangent and in the same

direction as the motion.

Vectors can be resolved into components as shown

in figure 3.

Instantaneous velocity, v :

vx = v cos

vy = v sin

The components separately follow the same laws of

motion as if the motion were in a straight line.

Figure 3

Time, t t

r

Figure 2

Displacement, r

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iv) Average acceleration, a

Average acceleration of an object is defined as the change in velocity, v divided by the

total time taken.

o

o

v v va

t t t

v) Instantaneous acceleration, a

Instantaneous acceleration is the acceleration of an object at particular instant of time.

Or it is also defined as the limit of the average acceleration as the time interval goes to

zero.

lim

0

va

t t

The instantaneous acceleration is the gradient of the tangential line at a particular time

drawn on the curve, as shown in figure 4.

The direction of the vector acceleration is the same direction as the motion.

2.6 EQUATIONS OF KINEMATICS IN 2-D FOR

CONSTANT ACCELERATION

The equations of motion in 2-D is same as before in chapter 2, only it is with subscripts for each

direction of motion.

x-component y-component

2

2 2

)

1)

2

1)

2

) 2

x ox x

ox x

ox x

x ox x

a v v a t

b x v t a t

c x v v t

d v v a x

2

2 2

)

1)

2

1)

2

) 2

y oy y

oy y

oy y

y oy y

a v v a t

b y v t a t

c y v v t

d v v a y

where:

Velocity, v

Time, t t

v

Figure 4

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x = the x (horizontal) displacement of the object

y = the y (vertical) displacement of the object

vox = the initial velocity of the object in the x-direction

voy = the initial velocity of the object in the y-direction

vx = the final velocity of the object in the x-direction

vy = the final velocity of the object in the y-direction

ax = the acceleration of the object in the x-direction

ay = the acceleration of the object in the y-direction

t= the elapsed time

EXAMPLE 2.1

A spacecraft is traveling with a velocity of vox = 5480 m/s along the x-direction. Two engines are

fired for 842s. The acceleration along the x-comp and y-comp are given as below:

Engine one: ax = 1.2 m/s2

Engine two: ay = 8.4 m/s2

Find final vx and vy?

Answer

Velocity in x-direction: vx = vox + axt

= 5480 +1.2(842)

= 6490 m/s

Velocity in y-direction: vy = voy + ayt

= 0 + 8.4(842)

= 7073 m/s

Final speed:

2 26490 2073x yv v v 9600m/s

2.7 PROJECTILE MOTION

Projectile motion is motion under the influence of gravity. If there were any other force besides

force of gravity acting upon an object, then that object would not be a projectile.

ASSUMPTIONS FOR PROJECTILE MOTION

The free-fall acceleration g is constant over the range of motion, and directed down

The effect of air friction is negligible

With these assumptions, an object in projectile motion will follow a parabolic trajectory

It can be divided into 3 TYPES:

a) HORIZONTAL projections

b) Projections with arbitrary angles (SAME LEVEL)

c) Projections with arbitrary angles (DIFFERENT LEVEL)

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i) HORIZONTAL PROJECTION

Example: The package falling from the plane is an example of horizontal projectile. (as shown

in Figure 5)

The above information can be summarized by the following table

x-component y-component

ax = 0 m/s2 ay = g = - 9.8m/s

2

vox = vx = constant

where ox

xv =

t

voy = 0 m/s (because object projected

in horizontal direction)

kinematics equations in y-comp:

2 2

2

)

) 2

1)

2

y oy

y oy

oy

i v v gt

ii v v gy

iii y v t gt

Instantaneous velocity at an instant of time: 2 2

x yv v v

vox

ymax

xmax

t = 0

vx

vy

vx

vy

v

v

a = -g

Figure 5

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Direction of instantaneous velocity:

1

tan

tan

y

x

y

x

v

v

v

v

EXAMPLE 2.2

A plane drops a package of emergency supplies to stranded explorers. The plane is travelling

horizontally at 40.0 m/s at 100 m above the ground. Find

a) where the package strikes the ground relative to the spot it was dropped and

b) the velocity and the direction of the package just before it hits the ground.

Answer

a)

212

212

2

100 0 ( 9.8)

100 4.9

4.5

oyy v t gt

t

t

t s

When t = 4.5s

40(4.5)

ox

xv

t

x

x = 180m

b)

voy = 0 vox = vx = 40m/s

y = -100m

x

vy

vx

v

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0 ( 9.8)(4.5)

44.1 / ( )

y oy

y

y

v v gt

v

v m s downward

2 2

2 240 ( 44.1)

tan

44.1tan

40

x y

y

x

v v v

v

v

v

v = 59.6m/s

θ = 47.79° (below the horizontal)

ii) PROJECTION WITH ARBITRARY ANGLE (SAME LEVEL)

Example: A football is kicked with initial speed of vo at an angle of above the ground. (as

shown in figure 6)

voy vo

vox a = -g

vx = v

Figure 6

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The above information can be summarized by the following table

x-component y-component

ax = 0 m/s2 ay = g = - 9.8m/s

2

vox = vo cos voy = vo sin

vox = vx = constant

where ox

xv =

t

vy = 0 m/s at maximum height

kinematics equations in y-comp:

2 2

2

)

) 2

1)

2

y oy

y oy

oy

i v v gt

ii v v gy

iii y v t gt

Instantaneous velocity at an instant of time:

2 2

x yv v v

Direction of instantaneous velocity:

1

tan

tan

y

x

y

x

v

v

v

v

Range, R

The range is the maximum horizontal distance traveled in a projectile motion. At the maximum

horizontal point, y = 0 and x = R 2 sin 2ov

Rg

Maximum height, H

Referring to figure 6, there is a peak of the curve. The height of this peak from the horizontal line

drawn from initial point of the trajectory is the maximum height. At the highest point, y = H and

x =½ R 2 2sin

2

ovH

g

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(EQUATION DERIVATION)

Y–component (Vertical)

Initial velocity , voy = vo sin

Acceleration ay = g = - 9.8m/s2

at maximum height (y = H) ,

vy = 0 m/s v2 = voy

2 + 2gy

02 = ( vo sin )

2 + 2gH

H = g

vo

2

sin 22

-- (1)

y = (vo sin )t + 2

1gt

2 --(4)

Let t = t1 = time taken to reach the highest point

(vy=0),

Using vy = voy + gt

0 = (vo sin ) + gt1

t1 = g

vo

sin -- (2)

Let t =T = time taken to return to the ground (y=0),

Using y = voyt + 2

1gt

2

0 = ( vo sin )T + 2

1gT

2

T = g

vo

sin2 -- (3)

From (2) T= 2t1

The vertical displacement after time t ;

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X–component (Horizontal)

- at horizontal motion object traveling with constant velocity vx ax=0

vox = vx = vo cos

ax = 0

The horizontal displacement after time t is :

Using vox = t

x x = voxt

x = (vo cos )t --(5)

The maximum horizontal displacement = Range = R,when t = T= time taken during return to

ground.

From (5) when x =R, t = T R = (vo cos )T

From (3) R = (vo cos ) (g

vo

sin2)

R = (g

vo

cossin22

)

R = (g

vo

2sin2

) --(6) * [ 2sin cos = sin 2 ]

When sin 2 = 1 , Range is maximum sin 90 = 1 = 45o

R = g

vo

2

(Maximum Range)

Symmetry:

The time for an object to reach its maximum height equals the time for it to return to its

original position. So, the total time of flight is twice of maximum time. (ttotal = 2tmax)

At a specific vertical point, the speed of an object is the same on its way up as on its way

down (however, its velocity is equal in magnitude but opposite in direction)

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EXAMPLE 2.3

A riffle fires a bullet with a speed of 250m/s at an angle of 37 above the horizontal.

a) Relative to the launch height, what maximum height does the bullet reach?

b) What is the horizontal range?

c) How long is the bullet in the air?

Answer

a) 2 2 2 2sin 250 sin 37

2 2(9.8)

ovH

g

H = 1155m

b) 2 2sin 2 250 sin 2(37)

9.8

ovR

g

R = 6130m

c)

cos

6130

250cos

ox

o

xv

t

xv

t

t

30.7s

iii) PROJECTION WITH ARBITRARY ANGLE (DIFFERENT LEVEL)

For an example, if we stand at the edge of the building’s roof and throw a ball up at an

angle,.

vo = 250m/s

vy

vx

37

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This is due to the matter that this kind of projection is a combination of horizontal

projections and projections with same level (refer figure 7).

But certain things have to be solved in different ways, for instance the time from launch

to maximum height will not be the same as the time from the maximum height to landing.

The symmetry that made this possible has gone.

EXAMPLE 2.4

A projectile is fired with an initial speed of 113 m/s at an angle of 60 o

above the horizontal from

the top of a cliff 49 m high. Find:

a) the time to reach the maximum height

b) the maximum height

c) the total time in the air

d) the horizontal range and

e) the components of the final velocity just before the projectile hits the ground.

Answer a) at max height, vy = 0m/s

Figure 7

113 m/s

y = -49 m

vo

600

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113sin 60

97.86 /

0 97.86 ( 9.8)

oy

oy

y oy

v

v m s

v v gt

t

t = 9.99s

b) 21

oy 2

212

y = v t + gt

y = 97.86(9.99) + (-9.8)(9.99)

y = 488.6m

Or we can used equation maximum height, H , which will give the same answer. 2 2sin

2

ovH

g

c) 21

2

212

2

2

2 2

49 97.86 ( 9.8)

4.9 97.86 49 0

., :

0.45

sin ,

o yy v t gt

t t

t t

solve the quadratic eq we will get

and t s

ce t is negative we reject solution of t

1t = 20.5s

d)

cos

113cos60

56.5 /

ox o

ox

ox

v v

v

v m s

( )

56.5(20.5)

oxx v t

x

x = 1158m

e)

97.86 ( 9.8)(20.5)

y oy

y

v v gt

v

x ox

y

v = v = 56.5m/s

v = -103m/s (dowaward)