Ch. 3 Vectors & Projectile Motion

Scalar Quantity

• Described by magnitude only– Quantity

• Examples: time, amount, speed, pressure, temperature

Vector Quantity

• Describe magnitude AND direction• Examples: velocity, force, acceleration,

resistance

• Vector quantities can be represented as “vectors” in physics diagrams– Arrow: points in direction of vector and specifies

the magnitude on top of the arrow10 m/s

Calculations using vectors

• Can be added or subtracted IF they are in the same plane

– Example: A boat travelling 10 m/s west encounters a down stream current, 5 m/s west. What will be the boats velocity travelling with this current?

Calculations using vectors

• Can be added or subtracted if they are in the same plane

– Example: A boat travelling 10 m/s west encounters a down stream current, 5 m/s west. What will be the boats velocity travelling with this current?

10 m/s + 5 m/s = 15 m/s

Journal #1

A plain travelling 50 m/s north encounters a head wind 10 m/s south. What is the resulting velocity of the plane?

Journal #1

A plain travelling 50 m/s north encounters a head wind 10 m/s south. What is the resulting velocity of the plane?

50 m/s – 10 m/s = 40 m/s

Journal #2: With the person sitting next to you…..

1. List what you remember about solving for the sides of a right triangle.

Write down as much as you know, be specific!!!

Vectors in different planes!!!!

• Consider a boat travelling North and a current in the water that is moving towards the east….

• We will solve this using Trigonometry!

Vectors in different planes!!!!

• Consider a boat travelling North and a current in the water that is moving towards the east….

• We will solve this using Trigonometry!

Vectors in different planes!!!!

• Consider a boat travelling North and a current in the water that is moving towards the east….

• We will solve this using Trigonometry!

this is the resultant (the path that the

boat will take)

a2+b2 = c2

a c

b

Important ConceptsSOH CAH TOA

Sin θ = opp Cos θ = adj Tan θ = opp hyp hyp adj

θHypotenuse

Adjacent

Opposite

Lets try a problem!!!

• A boat travelling with a velocity of 5 m/s North, encounters a current 2 m/s to the west.

• What is the resulting velocity?

– Draw a vector diagram – Fill in the knowns– Solve for the Resultant

Lets try a problem!!!

• A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west.

• What is the resulting velocity? 2 m/s

– Draw a vector diagram 5 m/s

Lets try a problem!!!

• A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west.

• What is the resulting velocity? 2 m/s

– Draw a vector diagram 5 m/s X

Lets try a problem!!!

• A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west.

• What is the resulting velocity? 2 m/s

( 2m/s)2 + (5 m/s)2 = X2

5 m/s X

Lets try a problem!!!

• A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west.

• What is the resulting velocity? 2 m/s

( 2m/s)2 + (5 m/s)2 = X2

4 + 25 = X2 5 m/s X√(29) = √ (X2)

5.39 = x

Lets try a problem on our own…Journal # 3

• A boat travelling at 9m/s south, encounters a current 16 m/s to the east.

• What is the resulting velocity and angle of displacement for the boat?

Lets try a problem on our own…Journal # 3

• A boat travelling at 9m/s south, encounters a current 16 m/s to the east.

• What is the resulting velocity and angle of displacement for the boat?

92 + 162 = X2 81 + 256 = X2 9 m/s

337 = X2 16 m/s

X = 18.35

Lets try a problem on our own…Journal # 3

• A boat travelling at 9m/s south, encounters a current 16 m/s to the east.

• What is the resulting velocity and angle of displacement for the boat?tan = opp

adj 9 m/s

tan = 16 16 m/s 9

= tan-1 (16/9) = 60

Components of Vectors

- Any vector can be “resolved” into its components.

- Its X and Y plane

Components of Vectors

- Any vector can be “resolved” into its components.

- Its X and Y plane - How???By creating a right triangle!!!

Components of Vectors

- Any vector can be “resolved” into its components.

- Its X and Y plane - How???By creating a right triangle!!!

Example….

What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?

What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?

6.4

37 °

What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?

Solve forX and Y Y

X

6.4

37 °

What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?

Solve for sin 37 = Y/6.4 cos 37 = X/6.4

X and Y 6.4 sin 37 = Y 6.4 cos 37 = X

Y Y = 3.85 X = 5.11

X

6.4

37 °

Projectile Motion

What Forces are acting on a Projectile?

• Initial Force that caused motion• Force of gravity

• Gravity causes the object to curve downward in a parabolic path (trajectory)

• An Object’s motion can be broken down into it’s horizontal and vertical component vectors.– x and y vectors

• Important Rule: Horizontal motion does NOT affect vertical

motion!!!!

Vx is constant and there is 0 acceleration!

Vy is changing and acceleration is due to gravity.

• At the top of a path, – there is no y velocity component– Vx component only!!!

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

V0

dy

dx

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

From Free Fall dy = ½ gt2

V0

dy

dx

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

From Free Fall dy = ½ gt2

From Linear Motiondx= v0t and that v0= dx / t

V0

dy

dx

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

v0 = 50 m/s t = 10 s dx= ?

Solving for the x-axis vector component• dx= v0t

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

Solving for the x-axis vector component• dx= v0t

• dx= (50 m/s)(10s)

• dx= 500 m

** Ch. 3 problem 41 in HW

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

But wait, t is unknown…..

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m• dx= 10 m • v0 is unknown

v0= dx / t

But wait, t is unknown….. And we can solve for it using dy = ½ gt2

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

dy = ½ gt2 OKAY – Solve for time

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

dy = ½ gt2 OKAY – Solve for time

50 m = ½ (10 m/s2) (t2) 10 s2 = t2 take the square root of both sides

t = 3.16 s

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

Knowing that t = 3.16 s, we can now solve for V0.

v0= dx / t = 10 m / 3.16 s = 3.16 m/s** problems 42 and 44 in the hw

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????• dy = ½ gt2

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????• dy = ½ gt2

• dy = ½ (10 m/s2) (0.2 s)2

• dy = 0.2 m ** problem 43

THE END!!!!