General Physics 1
(Phys 110 : Mechanics)
CHAPTER 4
Motion in 2D and 3D
Phys
110
Chapter 4 : Motion in 2D and 3D
Revision :
2. Displacement vector (∆𝒓):
1. Position vector (𝒓):
𝒓 𝒕 = 𝒙 𝒕 𝒊 + 𝒚 𝒕 𝒋 + 𝒛(𝒕)𝒌
Particle’s
motion in
2D
Position
vector 1
Lesson 4 of 5
Slide 1
Position
vector 2
Phys
110
Chapter 4 : Motion in 2D and 3D
Revision :
3. Average Velocity (𝒗𝒂𝒗𝒈 ):
direction of 𝒗𝒂𝒗𝒈 = direction of ∆𝒓
Lesson 4 of 5
Slide 2
4. Instantaneous Velocity (Velocity) (𝒗):
direction of 𝒗 : along
the tangent to path
Phys
110
Chapter 4 : Motion in 2D and 3D
Revision :
Lesson 4 of 5
Slide 3
6. Instantaneous Acceleration (Acceleration)
(𝒂):
5. Average Acceleration (𝒂𝒂𝒗𝒈 ):
direction of 𝒂 :
not related to path
Phys
110
Chapter 4 : Motion in 2D and 3D
Lesson 4 of 5
Slide 4
Objectives covered in this lesson :
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
to identify the launched angle of a projectile that is measured from the horizontal.
to resolve the initial velocity of the projectile into its components and write it in unit-vector notation.
to analyze the projectile motion into two one-dimensional independent motions: horizontal and vertical.
to identify the horizontal and vertical components of the acceleration of the projectile.
to calculate the horizontal and vertical components of the final velocity of the projectile after time t.
to calculate the horizontal and vertical displacement of the projectile after time t.
to calculate the maximum height that the projectile can reach.
to calculate the time that the projectile spend to reach any position.
to define the horizontal Range of the projectile.
to calculate the horizontal Range of the projectile.
Motion in 2D and 3D:
to calculate the maximum horizontal Range of the projectile.
Projectile Motion
to describe the path of the projectile (trajectory).
Phys
110
Chapter 4 : Motion in 2D and 3D
Lesson 4 of 5
Slide 5
4-5 Projectile Motion :
It’s a
2D
motion
Phys
110
Chapter 4 : Motion in 2D and 3D
4-5 Projectile Motion :
Q: can we consider the following as projectile motions:
1. a tennis ball in flight.
2. a plane in flight.
3. a duck in flight.
The images show
“Types of Projectile
Motions”.
Lesson 4 of 5
Slide 6
Phys
110
Chapter 4 : Motion in 2D and 3D
Lesson 4 of 5
Slide 7
4-5 Projectile Motion :
In studying the projectile motion in this course, we assume that “air” has no effect on
y
the projectile.
Max. height
Projectile
Projectile’s Path
“trajectory”
θ0
R
Launching
point
The range R
x
Landing
point
Phys
110
Lesson 4 of 5
Slide 8
Chapter 4 : Motion in 2D and 3D
4-5 Projectile Motion :
Initial Velocity (𝒗𝒐 ) :
Position vector 𝒓 of the motion
&
Velocity vector 𝒗 of the motion
Scalar components:
𝑣𝑜 is the magnitude of 𝑣𝑜 .
𝜃𝑜 is the angle between 𝑣𝑜 and the positive x direction.
change continuously
Acceleration vector 𝒂 of the motion
is constant and is always directed
downwards
Phys
110
Lesson 4 of 5
Slide 9
Chapter 4 : Motion in 2D and 3D
4-5 Projectile Motion :
Position vector 𝒓
Velocity vector 𝒗
y
y
vy
vx
y
y
x
v0y
y
x
v0x
y
v
vy
vx
v
v0
θ0
x
x
v
x
Phys
110
Chapter 4 : Motion in 2D and 3D
4-5 Projectile Motion :
Divide it into two 1D motions:
One in the x-axis (horizontal motion)
One in the y-axis (vertical motion)
and study them separately.
Lesson 4 of 5
Slide 10
Phys
110
Chapter 4 : Motion in 2D and 3D
Lesson 4 of 5
Slide 11
4-6 Projectile Motion Analyzed :
Projectile Motion
Horizontal Motion (x-axis)
Vertical Motion (y-axis)
No acceleration
(𝒂𝒙 = 𝟎)
Free-fall motion
(𝒂𝒚 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = −𝒈)
𝑣𝑥 = 𝑣𝑜𝑥
Substitute:
𝑥 − 𝑥𝑜 = 𝑣𝑜𝑥 𝑡
𝒂 → 𝒂𝒙 → (𝟎)
𝒗𝒐 → 𝒗𝒐𝒙 → 𝒗𝒐 𝒄𝒐𝒔𝜽𝒐
𝒗 → 𝒗𝒙
𝑣𝑜𝑥 = 𝑣𝑜 𝑐𝑜𝑠𝜃𝑜
Substitute:
𝒙 →𝒚
𝒂 → 𝒂𝒚 → (−𝒈)
𝒗𝒐 → 𝒗𝒐𝒚 → 𝒗𝒐 𝒔𝒊𝒏𝜽𝒐
𝒗 → 𝒗𝒚
Phys
110
Chapter 4 : Motion in 2D and 3D
4-6 Projectile Motion Analyzed :
Lesson 4 of 5
Slide 12
Phys
110
Lesson 4 of 5
Slide 13
Chapter 4 : Motion in 2D and 3D
4-6 Projectile Motion Analyzed :
the motion in x
substitute 𝒙𝒐 = 𝟎 , 𝒚𝒐 = 𝟎
Note:
When solving problems: always put the
origin of the xy-graph at the start of the
motion, so that 𝒙𝒐 = 𝟎 , 𝒚𝒐 = 𝟎.
the motion in y
then, combine by eliminating ( t )
Phys
110
Lesson 4 of 5
Slide 14
Chapter 4 : Motion in 2D and 3D
4-6 Projectile Motion Analyzed :
the motion in x
substitute 𝒙 − 𝒙𝒐 = 𝑹 , 𝒚 − 𝒚𝒐 = 𝟎
The Range (R): is the horizontal distance
the projectile has travelled when it returns
to its initial (launch) height.
the motion in y
then, combine by eliminating ( t )
and using 𝒔𝒊𝒏 𝟐𝜽 = 𝟐 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽
Phys
110
Lesson 4 of 5
Slide 15
Chapter 4 : Motion in 2D and 3D
4-6 Projectile Motion Analyzed :
Note: If the projectile’s final height is not the same as its initial height:
The horizontal range ≠ the horizontal distance travelled
The maximum horizontal range ≠ the maximum horizontal distance travelled
y
R
R
Horizontal
distance
x
Horizontal
distance
Phys
110
Lesson 4 of 5
Slide 16
Chapter 4 : Motion in 2D and 3D
4-6 Projectile Motion Analyzed :
Maximum Height:
Max. height
H
x
Phys
110
Chapter 4 : Motion in 2D and 3D
Lesson 4 of 5
Slide 17
4-6 Projectile Motion Analyzed :
Answer: (a) 𝑣𝑥 is constant.
(b) 𝑣𝑦 is initially positive, decreases to zero, and then becomes progressively more negative.
(c) 𝑎𝑥 = 0 throughout the motion.
(d) 𝑎𝑦 = −𝑔 throughout the motion.
Phys
110
Chapter 4 : Motion in 2D and 3D
Problem 21 :
Lesson 4 of 5
Slide 18
Phys
110
Chapter 4 : Motion in 2D and 3D
Problem 21 :
Lesson 4 of 5
Slide 19
Phys
110
Chapter 4 : Motion in 2D and 3D
Problem 38 :
Lesson 4 of 5
Slide 20
Phys
110
Chapter 4 : Motion in 2D and 3D
Problem 38 :
Lesson 4 of 5
Slide 21
Phys
110
Chapter 4 : Motion in 2D and 3D
Sample Problem (4-7) :
Lesson 4 of 5
Slide 22
Phys
110
Chapter 4 : Motion in 2D and 3D
Sample Problem (4-7) :
Lesson 4 of 5
Slide 23
Phys
110
Lesson 4 of 5
Slide 24 (last)
Chapter 4 : Vectors
Summary:
Motion in 2D and 3D:
Next lesson we will cover:
Projectile Motion.
Section (4-7).
Projectile Motion Analyzed.
Sample problem (4-10).
Any Questions?