ch. 3 vectors & projectile motion. scalar quantity described by magnitude only – quantity...
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![Page 1: Ch. 3 Vectors & Projectile Motion. Scalar Quantity Described by magnitude only – Quantity Examples: time, amount, speed, pressure, temperature](https://reader030.vdocuments.us/reader030/viewer/2022032414/56649eef5503460f94bffa6f/html5/thumbnails/1.jpg)
Ch. 3 Vectors & Projectile Motion
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Scalar Quantity
• Described by magnitude only– Quantity
• Examples: time, amount, speed, pressure, temperature
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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
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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?
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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
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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?
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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
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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!!!
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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!
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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!
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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)
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a2+b2 = c2
a c
b
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Important ConceptsSOH CAH TOA
Sin θ = opp Cos θ = adj Tan θ = opp hyp hyp adj
θHypotenuse
Adjacent
Opposite
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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
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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
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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
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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
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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
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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?
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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
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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
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Components of Vectors
- Any vector can be “resolved” into its components.
- Its X and Y plane
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Components of Vectors
- Any vector can be “resolved” into its components.
- Its X and Y plane - How???By creating a right triangle!!!
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Components of Vectors
- Any vector can be “resolved” into its components.
- Its X and Y plane - How???By creating a right triangle!!!
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Example….
What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?
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What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?
6.4
37 °
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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 °
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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 °
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Projectile Motion
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What Forces are acting on a Projectile?
• Initial Force that caused motion• Force of gravity
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• Gravity causes the object to curve downward in a parabolic path (trajectory)
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• An Object’s motion can be broken down into it’s horizontal and vertical component vectors.– x and y vectors
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• Important Rule: Horizontal motion does NOT affect vertical
motion!!!!
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Vx is constant and there is 0 acceleration!
Vy is changing and acceleration is due to gravity.
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• At the top of a path, – there is no y velocity component– Vx component only!!!
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Projectile Problem Solving
• Problems in which an object was dropped with a force in the x- axis
V0
dy
dx
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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
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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
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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?
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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
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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
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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?
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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…..
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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
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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
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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
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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
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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?
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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????
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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
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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
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THE END!!!!