Download Projectile Motion I. 2-Dimensional Motion • 2

Projectile Motion
I. 
2-Dimensional Motion
• 
2-dimensional motion simply means that the motion is caused by a force or forces that
have both an x and y component (remember vector addition?)
• 
The first situations we’ll study will involve two forces, one acting vertically and
one horizontally
• 
Projectile motion is influenced first by the force that projects the object, and then
only gravity, which pulls it vertically downward
• 
Circular motion consists of a force that pulls the object toward the center of motion, and
a force that wants the object to move in a straight line.
• 
A 2-dimensional force is exerted at an angle to the horizontal. It can be separated into
two perpendicular components that are independent of each other
ANY FORCE VECTOR EXERTED AT AN ANGLE TO THE HORIZONTAL WILL
HAVE A RESOLVABLE X AND Y COMPONENT OF FORCE
THE VECTOR SUM OF THESE COMPONENTS GIVES THE FORCE EXERTED
AT THE GIVEN ANGLE
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Vector Resolution Quick Review
Suppose you are pulling a box with a rope…

R
The force vector can be resolved into an
x and y component.
Ry

R

R
Rx
Ry
Rx
the vectors are related to each other
by a right triangle when the x and y
components are added head to tail
The force vectors Rx and Ry are components of the force vector R.
This force analysis says that a force applied at an angle gives a vertical
force and a horizontal force
A common feature of 2-D force problems will be to resolve the forces, make
a right triangle, and analyze the triangle.
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HOW TO WORK A PROBLEM
1. 
2. 
3. 
4. 
5. 
6. 
Read the problem CAREFULLY
Get a mental picture of the situation
Identify the variables you are solving for
DRAW A PICTURE
Analyze the picture with vectors, labeled parts, etc
Break out any parts of the picture you need such as vectors to
add, right triangles, etc
7. Break out any horizontal and vertical components and separate them
8. Decide which equations will help you solve the problem.
9.  Often, just looking at the UNITS will help you identify the equations
10. Show all of your work NEATLY
11. The answer must have the correct units or it is WRONG.
NOTICE THAT WE DON’T THINK ABOUT THE EQUATIONS FIRST.
WE THINK ABOUT THEM LAST.
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II. Projectile Motion
A. DEFINITIONS
• By DEFINITION, the MOTION of a PROJECTILE is influenced only by the force of
GRAVITY.
•  This is like free fall EXCEPT that PROJECTILE MOTION has a HORIZONTAL
direction and a VERTICAL direction
•  A projectile is NOT a rocket or any other object with its own source of thrust
•  The SHAPE of the path in projectile motion is always a PARABOLA (unless the projectile
is dropped straight down or thrown straight up)
•  Projectile motion is something a lot of people think they can describe, but their
understanding has misconceptions.
•  The most common misconception is that a moving object needs
a force to KEEP it in motion. We know this is NOT so (right?)
•  Another misconception involves the forces acting on it. After the
initial thrust or push to launch a projectile ONLY GRAVITY ACTS
ON IT (we’ll ignore air resistance)
•  Another misconception is the time it takes to hit the ground. People
think that the horizontal motion keeps the projectile in the air longer.
If I fire a projectile HORIZONTALLY (no angle up or down), it will hit
the ground at the same time as if I had just dropped the projectile
from the same initial height
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B. Characteristics of projectile motion
Projectile motion has horizontal and vertical components
horizontal motion
projectile
vertical motion
WHAT DO YOU THINK I’M TRYING TO TELL YOU WITH THIS DIAGRAM?
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horizontal projectile motion
Look at the vector lengths..all the
same.
vertical projectile motion
(aka FREE FALL)
Look at the vector lengths…
they get bigger.
Velocity increases,
object accelerates
Velocity is constant,
Acceleration is zero
Suppose we add the vertical and
horizontal components…place
them head to tail…the resultant
is the projectile motion of the
launched object
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SAY IT AGAIN…
in the absence of gravity,
horizontal motion at CONSTANT
VELOCITY (zero acceleration)
is the only motion path
If the projectile is pushed off the edge,
it only has a vertical motion path due
to gravity. It ACCELERATES downward.
with gravity, a vertical motion
component is added to the
horizontal component. BUT,
the vertical component
ACCELERATES down due
to gravity.
THE VERTICAL AND HORIZONTAL COMPONENTS ACT INDEPENDENTLY.
THE VERTICAL COMPONENT ACCELERATES, THE HORIZONTAL COMPONENT
DOES NOT
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REMEMBER. ONLY GRAVITY
IS INFLUENCING A LAUNCHED
PROJECTILE.
whether dropped or launched, these objects
hit the ground at the same time. If the object
is launched UP or DOWN, then this is no longer
true.
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C. Launching at an angle
Launching horizontally or at an angle…
Either way
THE HORIZONTAL VELOCITY DOES NOT CHANGE
THE VERTICAL VELOCITY CHANGES BY 9.8 m/s EVERY SECOND
images from the physics classroom on-line
horizontal
vertical
Look closely at the numbers
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Can you see that the vertical component of the motion (velocity) is the only one that
changes? It changes by -9.8 m/s every second. That is the acceleration due to gravity.
Can you also see that there is no vertical component when the projectile reaches its
MAXIMUM HEIGHT?
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I can tell from the picture that the cannon was fired with an initial velocity of

R
θ
19.6 m/s
R = 19.6 2 + 73.12 = 75.7
73.1 m/s
Also, the angle of the cannon must be
m
s
19.6
= 0.268
73.1
θ = tan −1 0.268 = 15.0 o
tan θ =
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D. TIME AND MAXIMUM HEIGHT
maximum height
ymax
x/2
X
t=0
launch
t = ttot
hit
ttot = 2t y max
The time at the projectile’s maximum height is half the total flight time
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E. 2-D KINEMATIC EQUATIONS (for projectiles launched at an angle)
Since the motion has 2-dimensions, the equations break out into a set of
horizontal and vertical equations. Otherwise, they are JUST LIKE the
1-D equations from the previous unit.
VERTICAL
HORIZONTAL
1 2
x = vix t + ax t
2
v fx = vix + ax t
1 2
y = viyt + gt
2
v fy = viy + gt
v 2fx = vix2 + 2ax x
v 2fy = viy2 + 2gy
x is horizontal displacement
since there is no horizontal acceleration, ax = 0 m/s2
y is vertical displacement
g is acceleration due to gravity
x = vix t
v fx = vix
v 2fx = vix2
these two are trivial. They just reinforce the idea that
horizontal velocity is constant
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Sample Problem…
A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees
with the horizontal. Determine the time of flight, the horizontal displacement, and
the peak height of the football.
25 m/s
ymax
45o
x and ttot
Break out some vectors
We want to know:
time of flight: ttot
horizontal displacement: x
peak height: ymax
25 m/s
vy
45o
vx
We can’t get horizontal distance without vx
We can’t get vertical height without vy
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25 m/s
vix = 25 cos 45 o
m
vix = 17.7
s
viy = 25 sin 45 o
viy
45o
vix
viy = 17.7
m
s
the initial direction of vix is +
the initial direction of viy is +
vfx = vix
vfy = -viy this is because of the shape
of a parabola. The direction changes after
the halfway point in the flight path
The horizontal distance only depends on vx and time
The vertical height depends on vy, g, and time
Time is missing from both solutions. Let’s try to solve for it first.
v fy = viy + gt
Solve for t
v fy = viy + gt
m
m
m
= 17.7 + (−9.8 2 )t
s
s
s
t = 3.6s
−17.7
Time of flight
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It’s EASY from here…
ttot = 3.6 s
x = vix t
m
x = (17.7 )3.6s
s
x = 63.7m
horizontal displacement:
peak height:
ttot = 2t y max
t y max
t y max
t y max
ttot
=
2
3.6s
=
2
= 1.8s
1 2
gt
2
m
1
m
y = (17.7 )1.8s + (−9.8 2 )1.8 2
s
2
s
y = 16.0m
y = viyt +
16