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```PHYSICS: Vectors and
Projectile Motion
Day 1
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.
A couple of quick definitions!
This should be review, but just in case…..
• Scalars
• Vectors
Remember: Force, Velocity,
Displacement, Acceleration are ALL
VECTORS!!!
Representing Vectors
• Because of direction, vectors do not add
the same way scalars do.
2 + 2 does not always equal 4!!!!
We will discuss 2 Ways to Add Vectors
1. Graphically (Today)
(Later this week)
• 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/note
s/forces/animations/AngleResultants.html
Another Reason why vectors are important:
http://www.stmary.ws/highschool/physics/home/notes/kine
matics/MotionTerms/animations/boatsRiver.swf
Find the Following Resultants
1.
5N
+
5N
= ?
3N
= ?
2.
+
6N
Components
• These two vectors are
known as components.
X - component
• In physics, we often
break vectors down into
horizontal (x) and vertical
(y) components.
Y - component
• Every vector can be
separated into two
equivalent vectors.
Calculating Components
• Use the Pythagorean Theorem to
calculate the missing component for each
velocity below.
a2 + b2 = c2
5 m/s
?
8 m/s
4 m/s
? 8.9 m/s
?
6.24 m/s
3 m/s
- 8 m/s
5 m/s
4 m/s
a2 + b2 = c2
(-8)2 + 42 = c2
64 + 16 = c2
80 = c2 c = 8.9 m/s
?
- 8 m/s
4 m/s
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.”
O
Sin (ø) =
H
H
O
ø
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
Cos (ø) =
H
H
A
ø
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
Tan (ø) =
A
O
A
ø
An easy way to remember…
• What should you do when you have dirty
feet?
»SOHCAHTOA
(pronounced: Soak A Toe.. Uhhh….)
SOH
Sine
Opposite
Hypotenuse
CAH
Cosine
Hypotenuse
TOA
Tangent
Opposite
What is a Function?
• 2x = Y
• What do I get for Y as I plug in numbers
for x?
Sin (x) Cos (x) and
Tan (x)
do the same thing!!!!
x
y
Plug in a number for the angle:
0
0
(x degrees)
1
2
and you get back a different number
(usually a decimal).
2
4
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
9
25o
r
sin (30) = x/20
cos (30) = y/20
sin (25) = 9/q
tan (25) = 9/r
.5 = x/20
.866 = y/20
.423 = 9/q
.466 = 9/r
(.5)20 = x
(.866)20 = y
.423q = 9
.466r = 9
x = 10
y = 17.32
q = 21.3
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 Homework
• Read Ch. 3 Section 3
• Answer Chapter Review Questions On
Pg. 109: 24, 25, 27, 29, 31, 32, 34
Trig.
Problems
#1
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.
HINT:
Draw a diagram showing the
information.
m
Trig.
Practice
#2
building. The foot of
from the building. The
14 feet on the building.
a. Find the length of the
foot.
Trig.
Practice
#2
b. What is the
angle between the
and the wall it leans
on?
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!
• 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 East
Plane Engines: 200 m/s North
2. Wind 30 m/s East
Plane Engines: 100 m/s North
3. Wind 120 m/s West
Plane Engines: 175 m/s North
Ch. 3.3 Projectile
Motion
Physics
Mr. Pacton
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
projectiles we ignore the
effects of air resistance
and/or wind.
The only force acting on a
projectile must be gravity!!!
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 Equations
Now lets look at an example!
Vx = constant
Vy = -gt
increasing
downward
Vx
*ONLY for horizontally
launched projectile.
Vy
1. So what is the horizontal velocity at 3 seconds for the cannon above?
Vx = constant = 100 m/s
2. What is the vertical velocity at 3 seconds for the cannon above?
Vy = -gt = -10x3 ~ -30 m/s
3. What is the total velocity at 3 seconds for the cannon above? (Hint: you
have to add the two vectors!!!! A2+B2=C2)
2
2
2
2
2
V = Vx + Vy = 100 + 30 = 10,000 + 900
2
V = 10,900 Now take Square root to find V = 104.4 m/s
Vocabulary: Range/Height
height
range
The range of a projectile is how far it travels
horizontally.
The height is how high up into the air it goes.
Projectile 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, Dx, is:
Dx = vixt
• Vertically, the Velocity increases downward with an
acceleration equal to g. The total vertical displacement,
Dy, is:
2
Dy = viyt – ½gt
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 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! 
Initial Velocity:
V
ø
Vx
Vix= V cos (ø)
Vy
Viy = V sin (ø)
And to Find V at
any time:
2
2
2
V = Vx + Vy
Velocity at any
time (t):
Vx = Vix = V cos (ø)
Vy = V sin (ø) - gt
Mathematical Summary
Breaking Down the Components
Horizontal
Vertical
0
g = -10 m/s 2
Velocity
constant
Vx = Vix
changing
Vy = Viy - gt
V2 = Vx2 + Vy2
Displacement
Dx = vixt
Dy = viyt – ½gt2
D2= x2 + y2
Acceleration
Sum
And
Vix = (Vi)(Cos q)
Viy = (Vi)(Sin q)
Quick Lab
• Look at Pg. 97 in your book.
• Complete the Quick Lab and answer all
questions. (Turn in 1 paper per table)
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.
```