This belongs to the Vertical Straight Line Motion of Kinematics.

1. TIPS FOR TAKING ONLINE CLASS
FOR YOURSELF
1: Be ready before the starting time
2: Treat an online course like a real course
3: Hold yourself accountable
4: Practice time management
5: Create a regular study space and stay organized
6: Eliminate distractions
7: Figure out how you learn best
8: Actively participate
9: Leverage your network
INSIDE THE CLASS
1: Open your camera
2: Mute your audio
3: take your pen, copy and other material sources
3: Don’t add your note on the screen
4: Do ask, write and share after the permission. It
will be provided at the end of the class
5: Be more disciplined.
1

2. This is a REGULAR CLASS. Please make a
note by writing down what you have
seen here.
Make a separate note for MECHANICS

3. UK Physics
“Nothing happens
until something
moves.”
― Albert Einstein
“He who thinks half-heartedly will not believe in God; but he who really
thinks has to believe in God.”
― Isaac Newton

4. UDAY KHANAL
Department of Physics
CCRC
4

5. TRY TO REMEMBER THE HORIZONTAL
STRAIGHTLINE MOTION!
I hope that you all did your assignment that
were provided in the previous class. If not,
please try to submit your work. I am checking
you all in the face book page. Thanks!
5

6. VERTICAL STRAIGHTLINE MOTION
Downward MOTION
These two are two possible
VERTICAL STRAIGHT LINE MOTION
Upward MOTION
1
2
6

7. g
DO YOU KNOW?
Acceleration due to gravity is
always in downward direction.
Proof
A ball dropped from a certain height
falls always vertically down without
any other external disturbance.
Why gravity has no horizontal effect?
Why gravity has no both effect?
7

8. Let’s FOCUS on EQUATION of KINEMATICS in case of
VERTICAL MOTION
2
2
1
2
2
t
t
t a
u
s
as
u
v
a
u
v
2





 They are valid for CONSTANT acceleration
Valid for both HORIZONTAL as well as VERTICAL motion
Each parameters are VECTORS except time.
In VERTICAL MOTION,
S – is appropriate to say height (h). It is from initial to final position. It is either up
or down. It depends up on the cases. It is a vector.
a – is appropriate to say g . It is always vertically down. It is a vector.
U – is either up or down. It depends up on the cases. It is a vector.
V – is either up or down. It depends up on the cases. It is a vector.
t – is time. It is a scalar.
GENERAL EQUATIONS OF MOTION
8

9. WHAT DID YOU SEE?
You saw that, only direction of g is
fixed. The direction of others can be
varied.
So we must use sign convention.
9

10. SIGN CONVENTION:
There is no hard and fast rule for the sign convention. It
depends up on the personal choice.
If you take UPWARD direction +ve then downward direction will be –ve
and vice versa.
A chooses B chooses
+ ve - ve - ve -+ve
They both are correct.
10

11. REMEMBER!
Sign convention never affect on
magnitude. They only changes the
sign.
11

12. QUIC TIPS
1: Identify the origin or initial point
from where motion starts.
2: Chose your appropriate sign
convention, either upward +ve or
downward +ve.
3: Identify each parameters whether
they are up or down.
4: And finally fit the suitable
equation.
12

14. LETS TRY TO FIND OUT THE ORIGIN or the point
from where the motion is started.
Ground
level
…………………………. origin
………………………………….
origin
………………………….
origin
………………………….
origin
Height of a pole
Are you clear about to find the origin?
Here, in each case you can choose your appropriate sign
convention as mentioned earlier before.
1 2 3 4
Ground
level
14

15. LET’S TRY TO IDENTYFY EACH PARAMETERS
&WRITE THE EQUATIONS PROPERLY IN DIFFERENT CASES.
A ball throwing vertically upward
Taking upward +ve
u
v
Origin or
initial point
…………….
g
h
initial
Final
v = u + (-g) t
v2 = u2 + 2 (-g) h
h = u t + ½ (-g) t2
If you try it by taking downward direction as +ve the you will get the
same expression (after taking negative sign if required.)
15

16. WRITE THE PROPER THREE EQUATIONS IN THE
FOLLOWING FOUR DIFFERENT CASES.
HOMEWORK
1
…………………….
Origin or
initial point
h
3
………………………….
Origin or
initial
point
………………………………….
Origin or
initial
point
2 4
………………………….
Origin or
initial
point
Final
point
Final point
………………………………………..
………………………………
Final
point
Final point
You can choose either direction as positive.
…… line represents
the returned path.
16

17. You MUST KNOW !
…………………..
.………………………
t t
Time of
ascent
Time of
descent
Vertical velocity is zero
Why vertical velocity decreases?
Is there any horizontal velocity?
In the absence of other effects except gravity,
Time of ASCENT = Time of DESCENT
Total time of FLIGHT = Time of ascent + Time of descent
= 2 x Time of ascent
or
= 2 x Time of descent.
The time up to
which a body
remains in the air
17

18. CALCULATION of TIME
Time of ASCENT only or Time of DESCENT only can be calculated by fitting any
one of the suitable equations having ‘t’ i.e.
2
2
1
at
ut
s
or
at
u
v




Any one equation can be used. It
depends up on the cases.
For TIME of ASCENT or TIME of DESCENT (t):
For TOTAL TIME OF FLIGHT (T):
Since, T = 2 x t
METHOD ONE
METHOD TWO
2
2
1
aT
uT
s 

18

19. REMEMBER!
10 m/s - 10 m/s
The ball hits the ground with the same magnitude of the
velocity with which it was thrown. Only the direction of
sign is negative.
Can you justify it? It is left for the
homework.
1 2
19

20. Which ball REACH the ground FIRST from A to
B? Can they reach in the same time?
Height of a pole
Ground
level
1
2
A
B
10 m/s
10 m/s
Hints: They take the same time.
Upward
Downward
?
20

21. NUMERICALS
21

22. Problem 1
A body is dropped from a tower of a height 100 m. Calculate,
(a) The time to reach the ground.
(b) The velocity with which the body strikes the ground. [g = 10m/s2 ]
100 m
origin
+ve
2
2
1
gt
ut
h 

drop
100m 0 m/s 10 m/s2
On solving, t = 4.5s
gh
u
v 2
2
2


On solving, v = 45 m/s
gt
u
v 

To find the velocity,
any one equation
can be used. Both
are suitable. On solving, v = 45 m/s
22

23. Problem 2
A ball is thrown vertically upwards with an initial speed of 20 m/s.
Calculate,
(a) The time taken to return to the thrower Total time of FLIGHT
(b) The maximum height reached. [g = 10m/s2 ] Hmax
u = 20m/s
………………………………
Hmax
V = 0 m/s
ta td
origin
+ve
0 20
10
Method 1
t
g
u
v )
(


For (a)
t = 2 s
Finally, T = 2 x t = 4 s
Method 2
2
)
(
2
1
T
g
uT
h 


0 20
10
T = 4 s
For (b)
max
2
2
)
(
2 H
g
u
v 


0 20
10
Hmax = 20 m
23

24. 8 m/s
steady
…………………….
8
m/s
----------------------------------------------------
Origin or initial point
Final point
+ve
Height
of
the
balloon
Released
point
2
)
(
2
1
t
g
ut
h 



h = 1005 m
Problem 3
A Sand bag is released from a balloon which is ascending with a steady
vertical velocity of 8 m/s. If the sand bag hits the ground 15 s later, what
was the height of the balloon above the ground when it was released.
24

25. 25

26. drop
----------------
origin
origin
60
m
Final point
x
60 - x
stone
20 m/s
ball
Ground
+ve
+ve
)
1
..(
..........
..........
10
2
1
0
)
(
2
1
2
2
t
x
t
g
ut
x




)
2
......(
..........
)
10
(
2
1
20
60
)
(
2
1
60
2
2
t
t
x
t
g
ut
x








t= 3 s
X = 45 m
Problem 4
A stone is dropped from the top of a tower 60m high. At the same time
another stone is projected vertically upwards from the ground with a
velocity of 20 m/s. Find when and where will they meet? [g = 10 m/s2 .
26

27. 2
20 m
¾v
1
drop
origin
+ve
h
g
u
v )
(
2
2
2


0 10 20
?
v = 20 m/s
To find the time interval you can use either
method 1 or method 2
Method 2
Ground
level
T = 3 s
0
2
)
(
2
1
)
( T
g
T
u
h 


¾ x 20
10
Striking point origin
+ve
Problem 5
A ball is dropped from a height of 20 m and rebounds with a velocity
which is ¾ of the velocity with which it hits the ground. What is the time
interval between the first and the second bounces? [g = 10 m/s2 ] .
27

28. drop
X = ?
+ ve
0.2 s
2.2 m
origin
A
B
2
)
(
2
1
t
g
ut
h 

2.2 10
?
u = 10 m/s
origin
+ ve
x
g
u
v )
(
2
2
2


10 0 10
x = 5 m
?
Problem 6
A ball is dropped from the cornice of a building takes 0.2 sec to pass a
window 2.2 m high. How far is the top of the window below the cornice?
[g = 10 m/s2 .
28

29. +ve
-
-
-
-
-
-
-
--
-
-
-
-
-
-
-
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Final point
2
)
(
2
1
t
g
ut
h 



75
m
?
75 10
u = 10 m/s
NOTE:
Here t is for whole path. U is
representing for whole path. g is
representing for whole path. h is also
representing for whole path.
Because h is a displacement. The
equal and opposite portion have
been canceled just. So each
parameter have been used for the
equal conditions.
Problem 7
A boy throws a ball nearly vertically upward from a point near the
cornice of a building. It just misses the cornice on the way down and hits
the ground 75 m below 5 sec after it leaves the boy’s hand. With what
velocity was the ball thrown? [g = 10 m/s2 .
29

30. 40 m
origin
+ve
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
-
-
-
-
-
-
-
--
-
-
-
-
-
-
-
2
)
(
2
1
t
g
ut
h 



?
40 9.8
u = 9.6 m/s2
t
g
u
v )
(



9.8
?
v = 29.6 m/s
Problem 8
A stone is thrown vertically upward from a point on a bridge located 40
m above the water. Knowing that it strikes the water 4s after releases,
determine
(a) The velocity with which the stone was thrown upward
(b) the velocity with which the stone strikes the water. [g = 9.8 m/s2 .
30

31. HOMEWORK
Problem 9
A boy standing on a long rail road car throws a ball straight upwards. The
car is moving on the horizontal road with an acceleration of 1 m/s2 and
projection velocity in the vertical velocity direction is 10 m/s. How far
behind the boy will the ball fall on the car? [Ans: 2 m]
Problem 10
A body falls freely from the top of a tower, and during the last second of
its fall, it falls through 25 m. Find the height of the tower. [Ans: 45 m]
Problem 11
A body dropped from the rest falls half of its total path in the last second
before it strikes the ground. From what height was it dropped? [Ans:
58.2 m]
31

32. Problem 12
A body is dropped from the top of a tower , 90 m high and at the same
time another stone is projected vertically upwards from the bottom. If
they meet half way up, find the velocity of the projected body. [Ans: 30
m/s]
Problem 13
A balloon is going up at a steady velocity of 20 m/s . When it was 60 m
above the ground, an object is dropped from it. How long does the
object take to reach the ground ? [Ans: 6 s]
32

33. SHORT QUESTIONS
33

34. 2 Can you have zero acceleration but a non
zero velocity? Explain with the help of graph.
1 Give an example of a case where velocity is
zero but acceleration is not zero.
…………
V = 0
g = 9.8 m/s2
Constant velocity
a = 0
34

45. The essence of SCIENCE: ask an impertinent question, and
you are on the way to a pertinent answer.

46. DID YOU ENJOYE THE CLASS?
Leave your valuable suggestions so that I will be
better for you all in the next class. Your
suggestions are highly appreciated.
NO?
Yes?