# Ch. 3 Vectors & Projectile Motion. Scalar Quantity Described by magnitude only – Quantity Examples: time, amount, speed, pressure, temperature.

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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 arrow 10 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)

a 2 +b 2 = c 2 a c b

Important Concepts SOH CAHTOA Sin θ = oppCos θ = adj Tan θ = opp hyp hyp adj θ Hypotenuse 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 = X 2 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 = X 2 4 + 25 = X 2 5 m/s X √(29) = √ (X 2) 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? 9 2 + 16 2 = X 2  81 + 256 = X 2 9 m/s  337 =  X 2 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?

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 X 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/6.4cos 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!!!!

V x is constant and there is 0 acceleration! V y is changing and acceleration is due to gravity.

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

Projectile Problem Solving Problems in which an object was dropped with a force in the x- axis V0V0 dydy dxdx

Projectile Problem Solving Problems in which an object was dropped with a force in the x- axis From Free Fall d y = ½ gt 2 V0V0 dydy dxdx

Projectile Problem Solving Problems in which an object was dropped with a force in the x- axis From Free Fall d y = ½ gt 2 From Linear Motion d x = v 0 t and that v 0 = d x / t V0V0 dydy dxdx

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?

v 0 = 50 m/s t = 10 s d x = ? Solving for the x-axis vector component d x = v 0 t

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 d x = v 0 t d x = (50 m/s)(10s) d x = 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?

d y = 50 m d x = 10 m v 0 is unknown v 0 = d x / 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? d y = 50 m d x = 10 m v 0 is unknown v 0 = d x / t But wait, t is unknown….. And we can solve for it using d y = ½ gt 2

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? d y = 50 m d x = 10 m v 0 is unknown v 0 = d x / t d y = ½ gt 2 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? d y = 50 m d x = 10 m v 0 is unknown v 0 = d x / t d y = ½ gt 2 OKAY – Solve for time 50 m = ½ (10 m/s 2 ) (t 2 ) 10 s 2 = t 2 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? d y = 50 m d x = 10 m v 0 is unknown v 0 = d x / t Knowing that t = 3.16 s, we can now solve for V 0. v 0 = d x / 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?

The only known given is time and we are determining the distance in the vertical direction d y. 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 d y. Which equation should we use???? d y = ½ gt 2

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 d y. Which equation should we use???? d y = ½ gt 2 d y = ½ (10 m/s 2 ) (0.2 s) 2 d y = 0.2 m ** problem 43

THE END!!!!

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