Projectile of Kinematics

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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. UDAY KHANAL
Department of Physics
CCRC
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4. PROJECTILE
Definition
A body projected without any fuel is
generally called a projectile. The
projected body moves only under the
effect of gravity alone.
Example: Bullet fired from a gun.
Can a motion of a plane be a projectile
motion?
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5. TRAJECTORY
It is the parabolic path taken by a
projectile.
trajectory
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6. X – Component Horizontal Straight-line motion
Y
–
component
Vertical
Straight-line
motion
Projectile motion
Y
X
Remember:
g
Horizontal component of V CONSTANT
Vertical component of V CHANGES

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Y
Vx
Vx Vx
Vx
Vx
Vy
Vy
Vy
Vy
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
H
g
. . . . . . . . . . .
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X
Y
U
P (x, y)
O Q
R
𝜃
𝒕 = 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒂𝒔𝒄𝒆𝒏𝒕 𝒕 = 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒅𝒆𝒔𝒄𝒆𝒏𝒕
T= 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒇𝒍𝒊𝒈𝒉𝒕
PROJECTILE FIRED AT AN ANGLE WITH THE
HORIZONTAL [Oblique Projection]
(0,0)

8. Consider a body is projected with an initial velocity ‘U’ at an angle ‘𝜃′ with
the horizontal.
Now,
Horizontal component of the initial velocity, 𝑈𝑥 = 𝑈𝑐𝑜𝑠𝜃
Vertical component of the initial velocity , 𝑈𝑦 = 𝑈𝑠𝑖𝑛𝜃
Here, horizontal component of the velocity is not affected by the
gravitational field (𝑔𝑥 = 0). So it remains constant through out the motion.
But, the vertical component of the velocity goes on decreasing at first and
becomes zero at certain point. Then the body moves down with increasing
velocity.
Let O(0,0) be the initial position and P(x,y) be the position of the projectile
after t sec.
Then,
𝑋 = 𝑈𝑥𝑡
𝑜𝑟 𝑋 = 𝑈𝑐𝑜𝑠𝜃 𝑡
→ 𝑡 =
𝑋
𝑈𝑐𝑜𝑠𝜃
And, Y = 𝑈𝑦𝑡 −
1
2
𝑔𝑡2
= 𝑈𝑠𝑖𝑛𝜃 𝑡 −
1
2
𝑔𝑡2
= 𝑈𝑠𝑖𝑛𝜃
𝑋
𝑈𝑐𝑜𝑠𝜃
−
1
2
𝑔
𝑋
𝑈𝑐𝑜𝑠𝜃
2
= 𝑋𝑇𝑎𝑛𝜃 −
1
2
𝑔
𝑈2𝑐𝑜𝑠2𝜃
𝑋2
∴ 𝐘 = 𝒂𝑿𝟐 − 𝒃𝑿𝟐: 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎
Hence the path of the projectile is parabolic in nature.
Maximum height (H): It is the greatest height to which
the projectile rises above the ground.
Here the suitable equation is ;
𝑉𝑦
2 = 𝑈𝑦2 − 2𝑔𝐻max
𝑜𝑟, 𝑈𝑦2 = 2𝑔𝐻𝑚𝑎𝑥
𝑜𝑟, 𝑯𝒎𝒂𝒙 =
𝑼𝒚
𝟐
𝟐𝒈
=
𝒖𝟐𝒔𝒊𝒏𝟐𝜽
𝟐𝒈
Total time of flight (T): It is the time taken by the
projectile to reach the same horizontal plane at Q.
Here the suitable equation is ;
ℎ = 𝑈𝑦𝑇 −
1
2
𝑔𝑇2
𝑜𝑟, 𝑈𝑦𝑇 =
1
2
𝑔𝑇2
𝑜𝑟, 𝑻 =
𝟐𝑼𝒚
𝒈
=
𝟐𝑼𝒔𝒊𝒏𝜽
𝒈
Horizontal Range (R): It is the horizontal
distance travelled by the projectile during the
total time of flight.
Here,
𝑹 = 𝑈𝑥𝑇 = 𝑈𝑐𝑜𝑠𝜃 ×
2𝑈𝑠𝑖𝑛𝜃
𝑔
=
𝑼𝟐𝒔𝒊𝒏𝟐𝜽
𝒈
Note:
For Rmax, 𝑠𝑖𝑛2𝜃 =
1 = sin90°
→ 2𝜃 = 90°
𝑜𝑟 𝜃 = 45°
Time of ascent or descent
under the affect of gravity
alone
𝑡 =
𝑇
2
=
𝑈𝑠𝑖𝑛𝜃
𝑔

9. 9
Note
We saw that R is maximum at 𝟒𝟓°

10. PROJECTILE FIRED HORIZONTALLY [Horizontal
Projection]
Consider a body is projected horizontally with an initial velocity ‘U
from the height h.
Now,
Horizontal component of the initial velocity, 𝑈𝑥 = 𝑈
Vertical component of the initial velocity , 𝑈𝑦 = 0
Here, horizontal component of the velocity is not affected by the
gravitational field (𝑔𝑥 = 0). So it remains constant through out the
motion.
But, the vertical component of the velocity goes on increasing.
Let O(0,0) be the initial position and P(x,y) be the position of the
projectile after t sec.
h
. . . . . . . . . . . . . . . . . . . . . . . .
- - - - - - - - - - - - - - - - - - - -
-
-
-
-
-
-
-
𝑉𝑥
𝑉𝑥
𝑉𝑦
𝑉𝑦
X
Y
𝛼
U
R
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
O
𝒕 = 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒅𝒆𝒔𝒄𝒆𝒏𝒕
(0,0)

11. 11
Then,
𝑋 = 𝑈𝑡
→ 𝑡 =
𝑋
𝑈
And, Y = 𝑈𝑦𝑡 +
1
2
𝑔𝑡2
=
1
2
𝑔𝑡2
=
1
2
𝑔
𝑋
𝑈
2
=
𝑔
2𝑢2
𝑋2
∴ 𝐘 = 𝒂𝑿𝟐 ∶ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎
𝑤ℎ𝑒𝑟𝑒, 𝑎 =
𝑔
2𝑢2
Hence the path of the projectile is parabolic in nature. Horizontal Range (R): It is the horizontal
distance travelled by the projectile during the
total time of descent.
Here,
𝑹 = 𝑈𝑡 = 𝑈 ×
𝟐𝒉
𝒈
Time of descent (t): It is the time taken by the projectile
to reach the ground.
Here the suitable equation is ;
ℎ = 𝑈𝑦𝑇 +
1
2
𝑔𝑡2
𝑜𝑟, ℎ =
1
2
𝑔𝑡2
𝑜𝑟, 𝒕 =
𝟐𝒉
𝒈
0
Magnitude of Velocity of the projectile at any instant:
Vx = U
Vy = Uyt + gt = gt = 𝑉𝑥
2 + 𝑉𝑦2 = 𝑈2 + 𝑔2𝑡2
→ 𝜶 = 𝑻𝒂𝒏
_𝟏 𝒈𝒕
𝑼
Direction: 𝑇𝑎𝑛𝛼 =
𝑉𝑦
𝑉𝑋
=
𝑔𝑡
𝑈

12. 12
Significance of projectile motion:

13. NUMERICALS
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14. VERTICAL
STRAIGHT
LINE
MOTION
HORIZONTAL STRAIGHT LINE MOTION
Y
X
HORIZONTAL
and VERTICAL
components are
INDEPENDENT to
each other. So
we treat them
separately
YOU MUST KNOW!
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X
Y
Vx
Vx Vx
Vx
Vx
Vy
Vy
Vy
Vy
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
𝑉 = 𝑉𝑥
2 + 𝑉𝑦2
H
g
. . . . . . . . . . .
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X
Y
U
P (x, y)
O Q
R
𝜃
𝒕 = 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒂𝒔𝒄𝒆𝒏𝒕 𝒕 = 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒅𝒆𝒔𝒄𝒆𝒏𝒕
T= 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒇𝒍𝒊𝒈𝒉𝒕
For Understand (Rough)

16. 16
Practical observation as for example:

17. EQUATION OF TRAJECTORY
2
2
2
2
,
cos
2
1
bx
ax
y
or
x
u
g
xTan
y






MAXIMUM HEIGHT
TOTAL TIME OF FLIGHT
g
u
T

sin
2
 T = 2 t
HORIZONTAL RANGE
g
u
R

2
sin
2

VELOCITY OF THE PROJECTILE AT ANY INSTANT
)
(
Tan
,
)
( 1
-
2
2
x
y
y
x
v
v
v
v 
 
range.
horizontal
same
the
achive
to
n
projectio
of
angles
possible
two
are
o
θ
o
90
and
o
θ
maximum.
be
will
range
horizontal
the
,
o
45
o
θ
At


17
g
u
H
2
sin2
2


Remember

18. 18

19. EQUATION OF TRAJECTORY
2
2
2
,
]
2
[
ax
y
or
x
u
g
y


MAXIMUM HEIGHT
TIME OF DESCENT
g
h
t
2

HORIZONTAL RANGE
g
h
u
H
2

VELOCITY OF THE PROJECTILE AT ANY INSTANT
)
(
Tan
,
)
( 1
-
2
2
x
y
y
x
v
v
v
v
v 

 
19
Remember:

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21. 21

22. 22

23. 23
Problem 1
A projectile is fired from the ground with a velocity of 300 m/s at 30o
with the horizontal. Find the horizontal range, greatest height and the
time to reach the ground. [7794m, 1125 m, 15 s ]
1125
2
sin
height,
Greatest
2
2


g
u
H

s
15
sin
2
ground,
reach the
to
Time


g
u
T

m
g
u
R
7794
2
sin
Range,
Horizontal
2




24. Problem 2
A batter hits a base ball so that it leaves the bat with an initial speed
37 m/s at angle of 53o . Find the position of the ball and the magnitude
and the direction of its velocity after 2 seconds. Treat the base ball as a
projectile.
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26. 26
Problem 3
A base ball is thrown towards a player with an initial velocity 20 m/s and
45o with the horizontal, at the moment the ball is thrown the player is
50 m from the thrower. AT what speed and in what direction must he
run to catch the ball at the same height at which it is released? [3.5 m/s,
toward thrower]
50 m
45o
20 m/s
R = ……
T = ……
d
T
d
v 
thrower.
the
towards
m/s,
5
.
3


T
d
v

27. 114 m
……………………………..
23o
23o
Origin
+ ve
−ℎ = 𝑢𝑦𝑡 + 1/2 −𝑔 𝑡2
Final point
R
114
90 m/s
𝑈𝑥 = 90 cos 230
𝑈
𝑦
=
90
sin
23
0
10
t = 9.45 (+ ve value)
R = Ux t = 782 m
Problem 4
An airplane is flying with a velocity of 90 m/s at an angle of 23o above
the horizontal. When the plane is 114 m directly above a dog that is
standing on level ground, a suitcase drops out of the luggage
compartment. How far from the dog will the suitcase land? Ignore the
air resistance.

28. 28
Problem 5
A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60o
with the horizontal. Calculate the horizontal range covered by the bullet. Also
calculate the maximum height attained. [865 m, 375m]
Use DIRECT Formula

29. 29
Problem 6
A stone is projected horizontally with 20 m/s from top of a tall building.
Calculate its position and the velocity after 3 s neglecting the air resistance.
[36 m/s, 56.3]

30. Problem 7
A projectile is lunched with an initial velocity of 30 m/s at an angle of 60o
above the horizontal. Calculate the magnitude and direction of its velocity 5 s
after launch. [28.3 m/s, 57o from horizontal ]
Problem 8
A projectile is fired from the ground level with velocity 150 m/s at 30o to the
horizontal. Find it’s horizontal range. What will be the least speed with which
it can be projected to achieve the same horizontal range. [1949 m, 140 m/s ]
Problem 9
An object is dropped from the top of the tower of height 156.8 m and at the
same time another object is thrown vertically upward with the velocity of
78.1 m/s from the foot of the tower, when and where the object meet? [ 2 s,
20 m below from top ]
30

31. Problem 10
A body is projected horizontally from the top of a tower 100 m high with the
velocity of 9.8 m/s . Find the velocity with which it hits the ground? [ 45.82
m/s ]
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Problem 11
Two tall buildings are 40 m apart. With what speed must a ball be thrown
horizontally from a window 120 m above the ground in one building so that
it will enter a window 40 m from the ground in the other. [ 10 m/s]
Problem 12
A body is projected downward with the speed 20 m/s at an angle of 30o with
the horizontal from the top of a building 40 m high.
(a) How long will it take to strike the ground? [ 2 s]
(b) How far from the foot of the building will it strike the ground? [34.6 m]
(c) With what velocity will it strike the ground? [ 34.9 m/s, 60o ]

32. 32
Problem 13
The angle of elevation of an anti-aircraft gun is 70o and the muzzle velocity is
900 m/s . For what time after firing should the fuse be set if the shell is to
explode at an altitude of 1500 m. [ 1.79 s or 170.8 s ]
Problem 14
Find the velocity and direction of projection of a shot which passes in a
horizontal direction just over the top of a building which is 50 m off and 25 m
height. [ 31.6 m/s , 45o ]

33. SHORT QUESTIONS
1: A ball is thrown with a velocity v at an angle
@ with the horizontal. Can an another angle of
projection to achieve the same horizontal
range? Justify.
o
o
-
90
and 
 o
Show the same expression.
2: A bomb is to be dropped from a moving
helicopter on a target on the ground. Explain
how it can hit the target? [draw fig]
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34. 5: Can the direction of velocity of a body be
changed when its acceleration is constant?
[Vertically up]
6: Can an object with constant acceleration
reverses its direction? Explain. [ Vertically up]
4: A projectile is moving on a parabolic path
without air resistance. Is there is any point in
which acceleration is perpendicular to the
velocity? Explain.[ Maximum height]
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35. 7: From a high tower one ball is dropped and
another is simultaneously projected
horizontally. Neglecting the air resistance,
which ball will reach the ground earlier? [
Same]
8: Find the angle of projection at which
horizontal range and the maximum height are
equal. [75.96o ]
35

36. 9: What would be the effect on maximum
range in doubling the initial velocity of a
projectile? [R’ =4R]
10: A projectile fired at angle of 18o has
certain horizontal range. State another angle
of projection for the same horizontal range. [
72o ]
36

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

38. 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.
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