PHYSICS: Vectors and PHYSICS: Vectors and Projectile MotionProjectile Motion

Today’s Goals

Students will:

1. Be able to describe the difference between a vector and a scalar.

2. Be able to draw and add vector’s graphically and use Pythagorean theorem when applicable.

Day 1

A couple of quick definitions!A couple of quick definitions!

This should be review, but just in case…..This should be review, but just in case…..

• Scalars

• Vectors

Remember: Remember: Force, Velocity, Force, Velocity, Displacement, Acceleration are ALL Displacement, Acceleration are ALL VECTORS!!!VECTORS!!!

Representing VectorsRepresenting Vectors• Because of direction, vectors do not add the

same way scalars do.

2 + 2 does not always equal 4!!!!2 + 2 does not always equal 4!!!!

We will discuss 2 Ways to Add Vectors1. Graphically (Today)

2. Component Addition (Later this week)

Adding VectorsAdding Vectors• When two vectors are added together, the

results are a vector called a RESULTANT.

• In order to find the resultant, you must draw the two vectors “head to tail.”

Example: http://www.stmary.ws/highschool/physics/home/notes/forces/animations/AngleResultants.html

Another Reason why vectors are important:http://www.stmary.ws/highschool/physics/home/notes/kinematics/MotionTerms/animations/boatsRiver.swf

Find the Following ResultantsFind the Following Resultants

1. 5 N

+ 5 N = ?

2.

+6 N 3 N = ?

ComponentsComponents• Every vector can be

separated into two equivalent vectors.

• These two vectors are known as components.

• In physics, we often break vectors down into horizontal (x) and vertical (y) components.

X - component Y -

co

mp

on

ent

Calculating Components

• Use the Pythagorean Theorem to calculate the missing component for each velocity below. a2 + b2 = c2

3 m/s

4 m/s

4 m/s

- 8 m/s

?

?

5 m/s

?

8 m/s5 m/s

8.9 m/s6.24 m/s

a2 + b2 = c2

(-8)2 + 42 = c2

64 + 16 = c2

80 = c2 c = 8.9 m/s

4 m/s

?

- 8 m/s

Right Triangle Trig.Right Triangle Trig.Q: WHAT ARE TRIG FUNCTIONS?A: Sin Cos Tan

• Three functions that are useful when dealing with right triangles.

• These functions are used to determine missing sides of right triangles when other information is known.

Sine (sin)

• The Sin of any angle is equal to the ratio of the length of the “opposite side” and the length of the “hypotenuse.”

Sin (ø) =HO

O

ø

H

Cosine (cos)

• The Cosine of any angle is equal to the ratio of the length of the “adjacent side” and the length of the “hypotenuse.”

A

ø

HCos (ø) =HA

Tangent (tan)

• The Tangent of any angle is equal to the ratio of the length of the “opposite side” and the length of the “adjacent side.”

O

A

ø

Tan (ø) =AO

An easy way to remember…An easy way to remember…

• What should you do when you have dirty feet?

»SOHCAHTOA

(pronounced: Soak A Toe.. Uhhh….)

SOH CAHTOASine

OppositeHypotenuse

CosineAdjacentHypotenuse

TangentOppositeAdjacent

What is a Function?What is a Function?

• 2x = Y

• What do I get for Y as I plug in numbers for x?

x y

0 0

1 2

2 4

Sin (x) Cos (x) and Tan (x)

do the same thing!!!!

Plug in a number for the angle:

(x degrees)

and you get back a different number (usually a decimal).

Next, following SOHCAHTOA, the number you get back from the function, is equal to O/H, A/H, or O/A!

Trig. Practice Problems

#1

a.  Find x.b.  Find y.

c.  Find q.d.  Find r.

q

r25o

9

sin (30) = x/20

.5 = x/20

(.5)20 = x

x = 10

cos (30) = y/20

.866 = y/20

(.866)20 = y

y = 17.32

sin (25) = 9/q

.423 = 9/q

.423q = 9

q = 21.3

tan (25) = 9/r

.466 = 9/r

.466r = 9

r = 19.3

Today’s Objective

By the end of today…

1.You must be able to find the magnitude and direction of a resultant mathematically.

Weekly HomeworkWeekly Homework

• Read Ch. 3 Section 3

• Answer Chapter Review Questions On

Pg. 109: 24, 25, 27, 29, 31, 32, 34

Trig. Problems

#1

HINT:  

Draw a diagram showing the information.

                

m

From a point on the ground 25 meters from the foot of a tree, the angle of elevation

of the top of the tree is 32º.  (Meaning you have to look up 32o to be looking at the top of the tree.) Find to the nearest meter, the height of

the tree.

                 

A ladder leans against a building. 

The foot of the ladder is 6 feet from

the building.  The ladder reaches

height of 14 feet on the building.

a.  Find the length of the ladder to the nearest foot.

Trig. Practice

#2

                 

b. What is the angle between the top of the ladder and the

wall it leans on?

Trig. Practice

#2

Trig. Practice Problem #3

3. A plane travels 25 km at an angle 35 degrees to the ground and then changes direction and travels 12.5 km at an angle of 22 degrees to the ground. What is the magnitude and direction of the total displacement?

Quiz!Quiz!• The Vectors listed below represent the velocity of an

airplane and the velocity of wind. Each Plane is trying to Go North. Draw a Vector Diagram and Calculate the magnitude and direction of the resulting velocity.

1.Wind: 50 m/s EastPlane Engines: 200 m/s North

2. Wind 30 m/s EastPlane Engines: 100 m/s North

3. Wind 120 m/s WestPlane Engines: 175 m/s North

Ch. 3.3 Projectile Ch. 3.3 Projectile MotionMotion

Physics

Mr. Pacton

What is a projectile?What is a projectile?• A projectile is an object upon which the

only force acting is gravity.

• There are a variety of examples of projectiles:

1. An object dropped from rest

2. An object which is thrown vertically or horizontally

3. An object which is thrown upwards at any angle

NOTE: When we talk about projectiles we ignore the effects of air resistance and/or wind.

The only force acting on a projectile must be gravity!!!

Breaking it up…Breaking it up…• It is extremely useful in physics to break

up this motion into vertical and horizontal parts.

Horizontal Motion Vertical Motion

No Force Affecting Motion

The force of gravity acts downward causing the projectile to “fall”

No Acceleration Downward acceleration "g" is at ~ 10 m/s2

Constant Velocity

Changing Velocity (increasing by ~10 m/s each second)

Projectile Motion EquationsProjectile Motion Equations

Now lets look at an example!

Vx = constant

Vy = -gt increasing downward

*ONLY for horizontally launched projectile.

Vx

Vy

1. So what is the horizontal velocity at 3 seconds for the cannon above?

2. What is the vertical velocity at 3 seconds for the cannon above?

3. What is the total velocity at 3 seconds for the cannon above? (Hint: you have to add the two vectors!!!! A2+B2=C2)

Vx = constant = 100 m/s

Vy = -gt = -10x3 ~ -30 m/s

V2 = Vx

2 + Vy

2 = 100

2 + 30

2 = 10,000 + 900

V2 = 10,900 Now take Square root to find V = 104.4 m/s

Vocabulary: Range/HeightVocabulary: Range/Height

The rangerange of a projectile is how far it travels horizontally.

The heightheight is how high up into the air it goes.

rangerange

heightheight

Projectile Motion EquationsProjectile Motion Equations

• We can also break down the displacement into Horizontal and Vertical parts! (x and y)

Projectile Motion Equations

• You know that Horizontally, the ball has a constant Velocity. So the total horizontal displacement, x, is:

x = vixt

• Vertically, the Velocity increases downward with an acceleration equal to g. The total vertical displacement, y, is:

y = viyt – ½gt2

Practice Problem

• A cannon ball is fired horizontally off a cliff at a velocity of 100 m/s. What is the magnitude and direction of the velocity vector after 5 seconds?

Projectile Motion EquationsProjectile Motion Equations

• Not every projectile is launched horizontally. • So the question remains, what general

equations can we apply to any projectile?» Now you need to combine and use your knowledge

of Trigonometry and Vector Math!

V

Vx

Vy

Initial Velocity:

Vix= V cos (ø)

Viy = V sin (ø)ø

Velocity at any time (t):

Vx = Vix = V cos (ø)

Vy = V sin (ø) - gtAnd to Find V at

any time:

V2 = Vx

2 + Vy

2

Mathematical SummaryMathematical Summary

constant changing

Vx = Vix Vy = Viy - gt

Displacement

Velocity

Acceleration 0

V2 = Vx2 + Vy

2

x = vixt y = viyt – ½gt2 D2= x2 + y2

g = -10 m/s2

Horizontal Vertical Sum

Breaking Down the Components

Viy = (Vi)(Sin )Vix = (Vi)(Cos )

And

Quick LabQuick Lab• Look at Pg. 97 in your book.• Complete the Quick Lab and answer all

questions. (Turn in 1 paper per table)• Additionally, Answer the following:

1. What is g, the acceleration due to gravity, and does it affect both projectiles equally?

2. Describe what you have learned about the vertical motion of projectiles. (Minimum 3 sentences.)

Practice Problem

• A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.

• Determine the horizontal and vertical components of the initial velocity.

• Determine the time that the water balloon is in the air.

• Determine the range of the water balloon.