EUREKA

PHYSICS The Ultimate Survival Guide

GRADE 12

Dr Megandhren Govender (Ph.D)

In everyday life we need to determine the position of objects. For example, the

position of a cell phone in a room, the location of an aircraft in flight, the position of

a cricket ball after being struck by the batsman, etc. Let us firstly consider the

motion of a particle along a straight line. Here by particle we could mean a soccer

ball, a car, a person or even a ship. This is really an approximation which is useful

when describing the motion of macroscopic objects at grade 12 level. We will assume

that we are working with SI units which means that length is measured in metres

and time in seconds. Coming back to the motion along a straight line - we choose as

our origin a point O on the line as well as a positive direction. If we consider East as

positive, then the displacement of the particle from the origin in an easterly

direction will be positive and on the opposite side of O, its displacement will be

negative. In general, if the particle is moving, its position along the line will be a

function of time.

Objects in the real world generally require three spatial dimensions to adequately

describe their motion through space. For example, a tumbling satellite in space may

follow a complicated path which requires three coordinates to fix its position in

space.

In this course we will simplify our investigations of motion by considering objects

moving along a straight line, ie., we restrict our motion to one dimension. To

describe the position of a particle as a function of time we require an origin (starting

point). Motion will be described with respect to this origin which may be at rest or

O positive displacement negative displacement

Chapter One - Mechanics

1.1 Frames of Reference

moving with constant velocity. For example a train leaves a station and travels

along a straight track for some time t and covers a distance Δx. We take the train

station to be the origin and we assign to the station the position x = 0. If the train

covers a distance x in time Δt, then it’s displacement with respect to the origin in

time Δt is Δx = x2 – x1 = x – 0 = x, east (say). Here we assumed that the

displacement of the train away from the station is positive. Should the train return

to the station, the displacement on its return journey will be negative. Upon arrival

at the station, its displacement will be zero. Its distance however, will not be zero.

Why?

The velocity of an object, just like its position, must be described relative to a

reference frame. For objects moving on the Earth’s surface, we take the Earth to be

a fixed frame. We measure position, velocity and acceleration of an object relative to

this fixed frame. We know that the Earth is orbiting the sun, and the sun together

with planets move through space. Someone in orbit above the Earth’s surface will

observe the motion of a particle on Earth differently to an observer standing on the

surface. Consider the following scenario:

Two cars are moving in the same direction along a freeway with speeds 100 km/hr

and 60 km/hr respectively. The driver of the faster car measures is speed as 100

km/hr relative to the Earth. The driver is at rest relative to his car. Good! What

about the relative speed between the two cars? The relative speed between the two

cars is 40 km/hr. Agreed? Let us slow our analysis a notch. Let

hrkmV

hrkmV

FE

SE

/100

/60

where VSE is the velocity of the slower car relative to the Earth and VFE is the

velocity the fast car relative to Earth. If we want the velocity of the fast car relative

to the slow car we write:

hrkm

VVV ESFEFS

/40)60(100

where VES = -VSE.

1.2 Relative velocity

ACTIVITY 1

An aircraft carrier moves smoothly across the water at a speed of VAW, relative to

the still water. A US marine walks along the flight deck to the front of the carrier

with velocity VMA, relative to the flight deck. The velocity of the marine relative to

the water is

VMW = VMA + VAW

We note that the resultant velocity of the marine relative to the water is the sum of

her velocity relative to the aircraft carrier and the velocity of the aircraft carrier

relative to the water. If the marine was walking in the opposite direction to the

motion of the aircraft carrier, then we subtract the velocities concerned. For

example, if the aircraft is moving at 50 km.hr-1 and the marine is walking at 7

km.hr-1 in the same direction as the carrier, then her velocity will be VMW = 50 + 7 =

57 km.hr-1 relative to the water. If the marine was walking in the opposite direction

then her velocity relative to the water will be

VMW = 50 - 7 = 43 km.hr-1.

Here we have taken the positive direction in the initial direction of the aircraft

carrier.

ACTIVITY 2

Another illustrative example is the following: You are driving along in your car

which is travelling at 100 km.hr-1. The speed has to be relative to something. This is

usually taken to be the Earth. The speed of the car is relative to the road. You the

passenger, is at rest relative to the car, but you are travelling at 100 km.hr-1

relative to the ground. Now suppose that while you are travelling at 100 km.hr-1,

another car pulls up alongside you, and that car is also travelling at 100 km.hr-1.

How fast are you travelling relative to the other car? The answer is 0 km.hr-1.

And then you may ask, what about a car travelling at 80km.hr-1 behind you?

Relative to the car behind you, you are travelling at 20km.hr-1, in the forward

direction.

DISCUSS EVERYDAY EXAMPLES OF RELATIVE VELOCITIES

1. Safe following distance while driving.

2. Navigation – air traffic control and sea-faring.

3. Use of conveyor belts in manufacturing, eg. bottling and canning of goods.

4. Escalators and treadmills.

5. Sport – skiing, biking, F1 racing, etc.

ACTIVITY 3

A man is driving a truck at 50 km.hr-1 along a straight flat road. He is spotted by a

motorcycle officer, who gives chase. By the time the truck driver sees the police

officer on his bike, the bike is travelling at 75 km.hr-1. What is the bike’s velocity

relative to the truck?

Let the velocity of the bike relative to the earth be VBE and the velocity of the truck

relative to the earth be VTE. We may write:

VBE = VBT + VTE

where VBT is the velocity of the bike relative to the truck.

75 = VBT + 50

VBT = 25 km.hr-1

Discuss the direction of VBT.

ACTIVITY 4

An escalator in a shopping mall moves at 1 m.s-1 and is 150 m long. If a girl

steps on at one end and walks 2 m.s-1 relative to the escalator, how much

time does she require to reach the opposite end if she walks

4.1 in the same direction as the escalator’s motion?

4.2 in the opposite direction?

SOLUTIONS

4.1 VgE = Vge + VeE

= 2 + 1 = 3m.s-1

Time taken to get across = D/VgE =150/3 = 50s.

4.2 Similarly, we obtain t = 150s.

OPTIONAL – 2-d RELATIVE VELOCITY

Here is a realistic problem involving relative velocities. A boat capable of travelling

at 4 m.s-1 in still water is used to cross a river which is flowing at 3 m.s-1. The river

is 1 km wide.

[a] If the captain of the boat wishes to travel directly across the river, at what

angle relative to the river bank should he head the boat?

[b] Should he head in the direction determined in [a] above, calculate the time

taken for the boat to reach the opposite bank.

[c] If the captain decides to head directly across the river, how far downstream

would he find himself, upon reaching the opposite bank?

SOLUTIONS

[a]

[b]

[c] kmXX

4

3

14

3tan

River bank

River bank

Vboat

Vriver

Vresultant

α θ

River bank

River bank

Vboat

Vriver

Vresultant

α

sV

Dt

smV tresul

2005

1000

/543 22

tan

090

4

3arcsin

4

3sin

boat

river

V

V

Wor

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k O

ne

1. Neo follows an exhaustive training program in preparation for

the Comrades marathon. One of her cardio-workouts requires

her spending twenty minutes on the treadmill. The treadmill

conveyor belt travels at 2 m.s-1. How fast must Neo run on the

treadmill for her to be stationary relative to the ground?

2. In the latest James Bond movie, agent 007 stands on the top

of a train which is travelling at a constant speed of 90km.hr-1

relative to the ground. A Russian agent is standing a distance

of 25 m away from agent 007, on the roof of the train. If 007

fires a dart at 25 m.s-1 towards the Russian agent, calculate

the time taken for the dart to strike the Russian agent.

Assume that the dart is fired in the opposite direction to the

train’s motion. Air friction is negligible. How will your answer

compare if the dart was fired in the same direction as the

motion of the train?

3. Nathi and Selvan decide to play a game of ping-pong aboard a

boat which is travelling at 30 m.s-1, relative to the water. They

stand 10 m apart. If Selvan hits the ball at 5 m.s-1 towards

Nathi (in the direction of motion of the boat), calculate the

time taken for the ball to reach Nathi. Will this time be the

same for spectators aboard the boat? What about for someone

watching the match from the pier?

4. A boat which is 10km from shore travels at a constant velocity

of 5m.s-1 towards the beach. A seagull aboard the boat starts to

fly at a constant speed of 10m.s-1 and heads for the shore.

Upon reaching the shore the seagull flies back to the boat and

repeats the journey to the shore. What distance does the

seagull cover when the boat finally reaches the beach? During

a test drive a BMW travelling at 100 km/hr in the left lane

attempts to overtake a (become even with) a VW Golf

travelling at 50 km/hr in the right lane. Calculate the time

Practice Problems

taken for BMW to line up with the Golf given front bumpers of the cars are

initially 100m apart

5. Two cars, standing a distance apart, start moving towards each other with

speeds 1 m/s and 2 m/s along a straight road. What is the speed with which

they approach each other?

6. If the cars in question 6 were initially 100m apart, how long will it take for

them to collide?

7. Two cars, standing a distance apart, start moving away from each other with

speeds 1 m/s and 2 m/s along a straight road. What is the speed with which

they move apart?

1. Neo will have to run 2 m.s-1 relative to the treadmill.

2. We note here that 007 and the Russian agent are at rest relative to each

other. The time taken for the dart to reach the Russian agent is independent

of the train’s velocity. The time taken for the dart to travel 25 m = D/V =

25/25 = 1s. The time will also be the same if the dart were fired in the same

direction as the train’s velocity. To understand we note that everything

happens on the train.

3. Similar to question 2. The time taken for the ball to travel between players

will be the same for the everyone aboard the boat, ie., time = D/V = 10/5 = 2s.

The situation is different for someone on the pier. If the ball is moving in the

same direction as the boat, then the velocity of the ball is added to the

velocity of the boat. If the velocities are in opposite directions, then the

resultant velocity is the difference between the two.

4. This is a classic question. A slight variation of this question was posed to

Richard Feynman, the renowned American physicist, when he was pursuing

SOLUTIONS

a PhD. The answer is quite simple. If we know how long the seagull is in the

air, then we can easily calculate the total distance covered by the seagull. The

boat approaches the shore at constant velocity. The time taken for the boat to

reach the shore must equal to the total flight time of the seagull = D/Vboat

=10000/5 = 2000s. Distance covered by the seagull = Vgull x t = 10 x 2000 =

20km. NICE!!!

5.

V12 = V1g + Vg2 = V1g + (-V2g) = 1 – (-2) = 3m/s

Note: V12 = velocity of car 1 relative to car 2

V1g = velocity of car 1 relative to ground.

V2g = velocity of car 2 relative to ground.

Vg2 = -V2g

V2 = -2m/s since the car is moving west, we chose east as positive.

6. t = D/V = 100/3 = 33.33 s

7.

V12 = V1g + Vg2 = V1g + (-V2g) = -1 – (2) = -3m/s or 3m/s westwards.

V1=1m/s V2=2m/s

Choose east as positive

V1=1m/s V2=2m/s

Choose east as positive

Consider an object being projected (thrown) vertically upwards with some initial

velocity vi . If we choose the upward direction as positive, then the acceleration of

the object is g = -9.8 m.s-2, since the acceleration due to gravity always points

towards the center of the Earth. It is also important to note that the object is

travelling in a straight line. If we ignore air resistance, then the equations required

to describe the object’s position, velocity and flight time are:

where y is the object’s vertical displacement, vi , its initial velocity, vf , the final

velocity, g, the acceleration due to gravity and Δt, the time interval in seconds.

NOTES

In the absence of air resistance:

1. At the highest point of its motion, the object’s final velocity is zero.

2. At the highest point, the object’s acceleration is g = 9.8m.s-2,

pointing downwards. We cannot switch off gravity!

3. The time taken to rise is equal to the time taken to fall back to the point

of projection.

AI

tvv

y

tgtvy

ygvv

tgvv

fi

i

if

if

2

)(

2

2

21

22

1.3 Vertical Projectile Motion

ON DUE TO GRAVITY - SPECIAL COMMENTS

In grade 11, you would have derived the following relationship:

2R

GMg

where G = 6.7 x 10-11 N.kg-2.m2, M is the mass of the gravitating body (for example

Earth) and R is the distance between the centres of the gravitating body and the

falling body.

ACTIVITY 1

Let us work out the acceleration due to gravity at the

Earth’s surface. Take the radius of the Earth to be

R = 6000 km and M = 6 x 1024kg.

In our derivation, we have made the following simplifications:

1. We ignored the rotation of the earth.

2. We have assumed that the mass Earths density is homogeneous.

Look up the meaning of homogeneous in your grade 10 chemistry notes.

3. We have also taken the Earth to be perfectly spherical.

A COMMON MISCONCEPTION

Most people believe that there is no gravity on the moon. This cannot be true since

the moon has mass. Also, gravity cannot be switched off.

DID YOU KNOW?

On *** July 2009 marked the 40th anniversary of man’s first landing on the moon.

********

sm

R

Gmg

E

EE

/81.9)106000(

)106)(107.6(23

2411

2

22

ACTIVITY 2

Using the fact that the moon’s mass is about 0.012 times the Earth’s mass and the

diameter of the moon is 3476 km, calculate the acceleration due to gravity on the

moon.

The acceleration due to gravity on the moon is given by

which is clearly not zero!

ACTIVITY 3 A ball is projected vertically upwards from the ground with an initial velocity of

10 m.s-1. Calculate

3.1 the maximum height reached by the ball above level ground.

3.2 the time taken for the ball to reach maximum height.

3.3 the speed of the ball after 1.5 s.

3.1 Take up as positive. Given vi = 10m.s-1, g = - 9.8 m.s-2.

At maximum height vf = 0m.s-1. We obtain

6

2

103476

)012.0(2

3

2

earthE

moon

moonmoon

gmG

R

Gmg

m

g

vvy

ygvv

if

if

2.108.9

100

2

2

22

22

22

3.2

3.3

The negative sign indicates that the ball is falling.

ACTIVITY 4

A stone is projected vertically upwards from a 50m high cliff, with an initial velocity

of 35 m.s-1. Ignore air resistance.

4.1 Calculate the time taken for the stone to reach its maximum height.

4.2 How fast is the stone travelling at t = 7 seconds?

4.3 How far is the stone above ground, after 8 seconds?

4.1 Given vi = 35 m.s-1, taking up as positive g = -9.8m.s-2. At the highest point

vf = 0 m.s-1.

4.2 Note that at t = 7s, the stone is travelling downwards. Do you agree?

sg

vvt

tgvv

if

if

02.18.9

100

1.7.4)5.1)(8.9(10 sm

tgvv if

1.6.33)7)(8.9(35 sm

tgvv if

sg

vvt

tgvv

if

if

57.38.9

350

The negative sign indicates that the stone is falling, as expected.

4.3

Height above ground = 50 - 33.6 = 16.4m

ACTIVITY 5

A cricket ball passes a window sill 15 m above a street with an instantaneous

velocity of 5m.s-1 upward. The ball was hit from the street below.

5.1 Define the acceleration of the body.

5.2 Determine the initial velocity of the ball immediately after it was hit.

5.3 What maximum height above the street level does the ball reach?

5.1 Acceleration is equal to the rate of change of velocity.

5.2 Given Δy = 15m, vf = 5m.s-1

, vi = ?

5.3 At maximum height, vf = 0m.s-1

m

tgtvy i

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)(

2

21

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21

12222

22

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)9.17(0

2

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Let us consider the example of a ball thrown straight up into the air which returns

to the thrower's hand after some time t. We are going to ignore air resistance. The

motion of the ball can be represented by a series of graphs which depict various

kinematical quantities as functions of time.

Before plotting our graphs, we will choose a POSITIVE direction for the motion of the ball. Our graphs will be plotted keeping this positive direction in mind. Since

the ball is thrown vertically upwards, we will choose up as positive. This means that

the acceleration of the ball on its way up as well as on its way down is negative. Do

you agree? Still confused? Recall that gravity always acts downwards (pulling every

mass towards the Earth’s center). We take up as positive, then down is surely

negative. If we plot the acceleration due to gravity as a function of time we obtain

the following graph for the whole motion (up and down) of the ball:

Note: 1. The area under the acceleration versus time graph yields the speed of the

object (velocity if direction is specified).

1.3.1 Graphs of Motion

(m.s-2

)

a

t (s)

2. The gradient of the acceleration versus time graphs gives us a quantity

called jerk (rate of change of acceleration).

Let us now plot the velocity of the ball as a function of time. The ball leaves the

thrower’s hand at a maximum velocity. At its maximum height, the velocity of the

ball is zero. The ball returns to the thrower’s hand with the same speed. Note that

the ball does not return to the thrower’s hand with the same initial velocity. The

initial velocity was upwards, the final velocity is downwards. You must agree that

the initial and final speeds are the same. Wow! Deep stuff!

Notes: 1. The area under a velocity versus time graph yields the displacement.

2. The slope of a velocity versus time graph gives the acceleration (in our

case of free-fall, the slope equals the acceleration due to gravity).

(m/s)

v

t

V0

-V0

Velocity is zero at maximum height

(s)

Slope = acceleration due to gravity

We will now plot the displacement of the ball as a function of time. Apart

from choosing up as positive, it is also important to choose your

reference or zero level. We will take our zero level to be ground level. It

is immediately obvious that when the ball strikes the ground, its total

displacement is zero as depicted by the graph below.

The instantaneous slope of the above graph gives the instantaneous velocity

over a short time interval. How do we find the slope of a curve? Besides

using Calculus, we can approximate the slope at each point by using

tangents.

At this time, the ball

returns to the ground

At this time, the ball

is projected

vertically upwards

from ground level.

We note that the gradient gradually diminishes as time varies, becoming zero

at the highest point. When the ball starts falling from the highest

point, the gradient is negative, but increases as expected.

The speed versus time graph is depicted below:

If we compare the speed versus time graph to the velocity versus time graph

we note that the speed does not take on negative values. This is

because speed is a scalar quantity.

(m/s)

v

t (s)

V0

Speed is zero at maximum height

(m)

t (s)

X

ACTIVITY 1

An object A, is thrown vertically upwards with a velocity of 20 m.s-1 and

subsequently falls back to ground, landing at 20m.s-1.

1.1 Sketch a velocity versus time graph representing the motion of object A.

1.2 Another object B, was thrown upwards 1s after object A was thrown, but

reached the ground at the SAME TIME as A. On the same graph, sketch the

motion of object B.

V

(m/s)

20

t(s)

A

B

1 2 3 4

ACTIVITY 2

2. Study the x t graph and answer the questions that follow.

2.1 Calculate the time taken to reach maximum height.

2.2 How long after launch does the object pass its launch point?

2.3 Calculate the initial velocity of the object.

2.4 Determine v (t) for the object.

2.5 Calculate the position of the object (with respect to ground) at t=3s.

2.1 2s 2.2 4 s

2.3

1

2

21

2

21

.8.19

)2)(8.9()2(1030

smv

v

gttvy

i

i

i

2.4 v(t) = 19.8 + gt

2.5 25 m

In the presence of air resistance there are three forces acting on the object as

depicted below:

If the sphere is moving slowly, the drag force will be small, as will the buoyancy

force, so that the sphere will be accelerated downwards by its weight. As the

velocity of the sphere increases, so does the drag force, so that at some point, the

buoyancy force and the drag force will balance the weight, and the sphere will cease

to accelerate downwards, i.e. it will reach a terminal velocity, vT . At this point,

This only applies to slow moving objects in very viscous fluids. It is found that for

slow moving spheres through a viscous medium, the drag force is proportional to the

square of the velocity of the sphere.

DID YOU KNOW?

A raindrop about the size of a fly falls at a rate of 9 m/s while a drizzle drops about

the size of a grain of salt falls at a rate of 2 m/s.

A human being tumbling through the air, without a parachute falls at a rate of

56 m/s. That’s fast!

BD FFW

FD = drag force

FB = buoyancy force

W = weight of object

1.3.2 Terminal Velocity

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k Boo

k Tw

o

1. A hot air balloon is rising vertically upwards with a constant

velocity of 8 m.s-1 when one of the passengers accidently lets

go of a cell phone while leaning over the edge. The cell phone

reaches the ground after 6 s.

Refer to the graph and answer the questions that follow.

1.1 What is the initial velocity of the cell phone?

1.2 What is the value of the slope of the graph? Explain.

1.3 Calculate the time x on the time-axis.

1.4 Which point on the path of the cell phone corresponds

to time x?

1.5 Calculate the magnitude of the velocity with which

the cell phone reaches the ground.

1.6 Calculate the height of the balloon above the ground

at the moment the passenger let’s go of the cell phone.

v (m/s)

0 x 6

t(s)

Practice Problems

2. Consider an experiment which resembles an experiment Galileo

performed from the leaning tower of Pisa. Ball A is dropped from rest from

a height of 80m. Ignore air resistance.

2.1 Calculate the time taken for A to reach the ground.

2.2 Ball B is thrown down from the same height 1.5 s after ball

A is released. Calculate the magnitude of the velocity with which

B must be thrown downwards in order to reach the ground at the

same instant as A.

3. A ball is dropped into a dark well; the sound of the ball hitting the bottom of the well

comes back after 2 s. Taking the speed of sound c = 340 m.s-1

, determine the depth of

the well.

4. Ball A is thrown vertically upwards with an initial velocity of 10 m.s-1. A

second ball, B is thrown vertically upwards, 5 seconds later, with an

initial velocity of 20 m.s-1.

4.1 Calculate the height above the ground at which the two balls will

meet?

4.2 When the two balls meet, their displacements will be the same,

however, they cover different distances. Explain.

5. Killer whales have been observed to routinely jump 7.2m above the water

surface. Calculate the speed of the whales as they leave the water.

6. A parachutist falling at a constant speed of 10 m.s-1 drops a cell phone when

she is 50m above the ground.

6.1 Calculate the time taken for the cell phone to hit the ground.

6.2 With what velocity will the cell phone strike the ground?

7. Melusi throws a ball vertically upwards to his friend who stands on the first

floor balcony which is 4m above. The ball is caught 1.5 s later by Melusi’s

friend.

7.1 Calculate the initial velocity of the ball.

7.2 What was the velocity of the ball just before it was caught?

8. A mountaineer stands on top of a 50m high cliff that overhangs a calm pool

of water. She throws two balls vertically downward 1 s apart and observes

that they strike the water at the same time. The first ball has an initial

velocity of 2m.s-1.

8.1 How long after the first ball is thrown will the two balls strike the

water?

8.2 Calculate the initial velocity of the second stone.

8.3 Calculate the velocity of each ball just before it strikes the water.

9. A ball is thrown vertically upwards from the ground with an initial

velocity of 25 m.s-1, at the same time, a ball is dropped from a building

that is 15m tall. After how long will the balls be at the same height above

the ground?

1.1 Take up as positive.

8 m.s-1

, upwards

1.2 slope = -9.8m.s-2

1.3 vf = vi + gt which gives t = 0.8 s

1.4 Point x corresponds to the maximum height reached by the cell phone from where its

starts falling towards earth.

1.8.50

)6)(8.9(8

sm

gtvv if

or vf = 50.8 m.s-1

downwards.

1.5 Time taken to fall the distance h, above the ground = 6 – 2(0.8) = 4.4s.

m

gttvy i

130)4.4)(8.9()4.4)(8( 2

21

2

21

Solutions

2.1

sg

yt

gt

gttvy i

04.48.9

)80(22

0 2

21

2

21

2.2 For ball B, the flight time will be t = 4.04 – 1.5 = 2.54 s

1

2

212

21

2

21

.05.19

54.2

)54.2)(8.9(80

sm

t

gtyv

gttvy

i

i

3. Think carefully here. The time t = 2s is made up of two parts – (i) the time taken for the

ball to travel down the well and (ii) the time taken for the sound signal to travel up the

well.

c

h

g

h22 ----- (1)

Note that we have a quadratic in h. Let h = k2, and so we can write (1) as

02

2

2

2

ckk

kkc

where we have defined g

c2

. Substitute the values for c and ω and solve for k or

equivalently h.

4.1 Taking the ground as our zero level and up as positive, we write the displacement equations

for both stones:

2

21

2

21

)5()5(20

10

tgtS

gttS

B

A

At the point of collision

mS

st

ggtt

SS BA

94.31)77.3)(8.9()77.3(10

39.8

02002040010

2

21

ie., the balls will collide 31.94 m above the ground.

4.2 That is correct. The ball that was thrown earlier would have travelled further. The two

balls collide when the first ball is falling while the second ball is moving upwards. At the

point of collision, both balls will be at the same height above the ground ie., their

displacements (the shortest (path that includes the start and end points) are equal.

5. That is surely an amazing height attained by these huge whales. The world record for

pole vaulting is *************. Anyway, let us work the velocity at which these whales

leave the water. At maximum height above the water’s surface, the velocity of the whale

is zero. Hence we write

6.1 It is important to note that the initial velocity of the cell phone is 10m.s-1

. Happy? Now it

is a simple matter of using the correct formula.

Note that we have two solutions. The solution that is physically viable is the positive time

value.

6.2 Here we required to calculate the final velocity of the cell phone, just before striking the

ground.

1

2

22

22

.9.1112.141

12.141)2.7)(8.9(20

2

2

smv

ygvv

ygvv

i

fi

if

st

t

t

tt

tt

tgtvy i

33.2 Choose

8.9

9.3210

)9.4(2

)50)(9.4(410010

05010)(9.4

))(8.9(2

11050

)(2

1

2

2

2

1.8.32)33.2)(8.9(10 sm

tgvv if

Note that in our calculations we have chosen the downward direction as positive.

7.. We take up as positive.

It is important to note that when the ball is caught, its velocity is not zero. We are given

that it takes 1.5s to travel a vertical distance of 4m.

7.1

7.2 )5.1)(8.9(10

tgvv if

&&&&&&&&&&&&&&&&&&&&&&&&

Δy = 4m

1

2

2

2

.105.1

)5.1)(8.9(2

14

)(2

1

)(2

1

sm

t

tgy

v

tgtvy

i

i

8. This is a lovely problem. We take down as positive.

8.1 We are really asked to calculate the time taken for the first ball to strike the water below.

A B

vi=2m.s-1

Δy=50m

st

t

t

tt

tt

tgtvy i

3 Choose

8.9

4.312

)9.4(2

)50)(9.4(442

0502)(9.4

))(8.9(2

1250

)(2

1

2

2

2

8.2 Since the second ball was released 1 s after the first ball was thrown, it means that it took

2 s for the second ball to strike the water below. Do you get this?

8.3 For ball A, we can write: For ball B, we can write:

9..

1

2

2

2

.2.152

)2)(8.9(2

150

)(2

1

)(2

1

sm

t

tgy

v

tgtvy

i

i

1.33.3123

)50(2

2

2

sm

vt

yv

tvv

y

if

fi

1.8.342.152

)50(2

2

2

sm

vt

yv

tvv

y

if

fi

A

B

Δy=15m

viA=25m.s-1

viB=0m.s-1

+ve

We now write the displacement equation for each ball.

2

2

2

2

)(9.4

)(2

1)(

)(9.425

)(2

1)(

t

tgtvy

tt

tgtvy

iBB

iAA

If we assume that ball A has travelled a distance x before colliding with ball B, then ball be

would have travelled a distance 15 – x. So we can write:

st

t

tttand

tx

xt

tgtvy

ttx

tgtvy

iBB

iAA

5

3

01525

)(9.415)(9.425)2( )1(

)2()(9.415

15)(9.4

)(2

1)(

)1()(9.425

)(2

1)(

22

2

2

2

2

2

Beautiful!