2-Dimensional Motion
Part I: Projectile Motion
Part II: Circular Motion
First of All, What is 2-D
Motion?
Before, we talked about motion in one
dimension (just the x axis)
Now we are investigating things that can
move it 2 dimensions (x and y)!
1D Motion
x
y
2D Motion
x
2-D Motion Part I:
Projectile Motion
A “projectile” is an object on which
the only force acting is gravity.
– These are things that fly through
the air!
(Examples: footballs, baseballs,
cars flying off cliffs, bullets,
cannon shells, and batman)
– Usually they are thrown/fired/etc.
but are not things like rockets,
which still have something
“pushing” them as they fly
through the air!
As always, gravity will accelerate the
object down.
The Dimensions are
Independent
Since 2D motion is
complicated, it is helpful
to think of the x and y
dimensions separately
when we analyze motion.
This is because the x and
y dimensions are
completely independent.
This lets us use all of our
favorite 1D equations
when solving problems!
Notice that after this idiot
jumps off the cliff his
velocity in the x dimension
does not change.
BUT……His velocity in the
y dimension is increasing
due to gravity.
The same is true for this
cannonball:
Gravity
What causes projectiles to
accelerate?
– Gravity!
In which dimension does
gravity exist?
– The vertical (y)!
The acceleration is always
9.81 m/s2 downward. This is
called “acceleration due to
gravity” or “g”
Since acceleration only
exists in the y dimension
this means it is always zero
for the x dimension.
Horizontal Velocity
Since we have established that acceleration only exists
in the y dimension and the acceleration in the x
dimension is zero, what does this mean about the
velocity of a projectile in the x dimension?
– It is always constant!
This needs to be remembered (by YOU)!
Solving Problems
We are going to take some time to go over
the problem solving process for these
projectile motion problems.
It is a “modified” version of GUESS.
The best part is that you don’t have to
learn any complicated equations… we will
use the ones you
already know from
the last unit! It’s easy!
Givens
Of course, the first step to solving
any problem is to list your givens.
Always begin by drawing a
diagram representing the motion.
And remember to include your
arrows for initial velocity,
acceleration, displacement, and
the positive direction.
Remember, we are separating the
variables into the two dimensions.
You will write out the same givens
table as before, but this time it will
be a different set for each
dimension (x or y)!
+
a
+
v
v
d
d
x
vo =
v=
ax =
t=
x=
y
vo =
v=
ay =
t=
y=
Equations
The equations you will use are the same
ones.
You will just be using the variables from
one dimension (x or y) at a time.
v f  v0  at
x  x0  v0t  12 at 2
2a( x  x0 )  v  v
2
f
2
0
Time
In projectile motion, we
will treat the x and y
variables as completely
different sets.
This means the givens
we use for one set cannot
be used with the other!
However, Time is a
variable that is always
the same for both x and
y.
x
vo = 25 m/s
v = 25 m/s
a = 0 m/s2
t = 4.0 s
x = 190 m
y
vo = 0 m/s
v = 48 m/s
ay = 10.0m/s2
t = 4.0 s
y = 22 m
Sample Problem
A cannon is placed at the edge of a cliff so the barrel is 60 meters above
the ground. The cannon fires a projectile horizontally with a velocity of
95 m/s. How far from the base of the cliff does the projectile land?
As always, we will start by drawing a
diagram of the situation. Remember
to label any givens and include a
direction vector for v, d, a, and +
+
Vx=95m/s
vy
ay
y=
60 m
+
x = ?? m
After you have drawn your picture, fill
in your Givens table.
And of course our unknown variable
is x. Always circle the unknown in the
table.
x
vo =95m/s
v = 95m/s
a = 0m/s2
t= ?
x= ?
y
vo = 0m/s
v= ?
ay = 10.0m/s2
t=?
y = 60 m
Sample Problem
A cannon is placed at the edge of a cliff so the barrel is 60 meters above
the ground. The cannon fires a projectile horizontally with a velocity of
95 m/s. How far from the base of the cliff does the projectile land?
+
Vx=95m/s
vy
dy=
60 m
ay
+
Now determine the equations you will
need to solve the problem.
Remember, we will keep the sets of
variables completely separate (except
time)!!!
dx = ?? m
x
vo =95m/s
v = 95m/s
a = 0m/s2
? s
t = 3.5
x= ?
y
vo = 0m/s
v= ?
ay = 10.0m/s2
t = ?3.5 s
y = 60 m
x = vot + ½ at2
This 2is the only equation
x = (95)(3.5) + ½ (0)(3.5)
we can use to solve for
x = 332.5 m
x. But we are missing
time! We need to use the
variables in the ydimension to find time
first.
y = vot + ½ at2
t = √(2y / a)
t = √[(2)(60) / (10.0)] = 3.5 s
Practice Problem
Thelma and Louise drive their car off
the edge of the grand canyon at a
speed of 30 m/s and die 3.4 seconds
later.
(a) how tall is the canyon at that spot?
Answer:
57.8 m
(b) how far away from the bottom of
the canyon do they land?
Answer:
102 m
Which of the following does not
effect the hang time of a
projectile?
A. Angled fired
B. Vertical velocity
C. Horizontal velocity
D. Height it was fired
at.
Angled Projectile Motion
Occasionally, you will
encounter problems
where the projectile is
initially fired at an angle.
– This means that the initial
velocity will have a
horizontal and a vertical
component.
You will need to calculate
these components in
order to fill in your givens
table on each problem.
Vector Components
In order to find the
components of a
vector (like velocity)
you will need to use
those timeless
Trigonometric
Functions.
Vector Components
So we have a
projectile that is fired
at an angle of 30o
with respect to the
horizontal at a
velocity of 50 m/s.
We need to think of
it like the projectile
is fired vertically and
horizontally at the
same time, giving it
both components.
vy
Θ=30o
vx
Vector Components
In order to calculate the
components, we need
to shift vy to make a
right triangle.
Then we can use trig
functions to solve for vy
and vx like they are
sides of a right triangle.
To solve for vx, we will
use cosine because it is
adjacent and we have
the hypotenuse.
vy
Θ=30o
vx
Vector Components
To solve for vx, we will use cosine
because it is the adjacent side and
we have the hypotenuse.
adj
hyp
v
cos   x
V
v x  V cos 
cos  
To solve for vy, use the same
process but with sine.
sin θ 
sin θ 
opp
hyp
vy
V
v y  V sin θ
vy
Θ=30o
vx
Sample Problem
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of
20m/s. What is the range of the soccer ball? (how far away does it land)
As always, start by drawing a
We
need toincluding
use trig functions
to
diagram,
all vectors.
solveasfor
vix and
viy your
And
always
circle
Beginunknown
to fill in what
you can
variable.
on your givens table
x
y
?
?
vi = 6.84m/s
vi = +18.8m/s
vf = 6.84m/s
?
vf = -18.8m/s
?
2
ax = 0 m/s ay = -10.0 m/s2
t= ?
t= ?
dx = ?
dy = 0 m
a
+
vix = VcosΘ
vix=(20)cos(70)
vix=6.84 m/s
vy
Θ=70o
vx
dx
viy = VsinΘ
viy=(20)sin(70)
viy=18.8 m/s
+
Sample Problem
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of
20.0m/s. What is the range of the soccer ball? (how far away does it land)
x
y
vi = 6.84 m/s
vf = 6.84 m/s
ax = 0 m/s2
t=?
dx = ?
vi = +18.8 m/s
vf = -18.8 m/s
ay = -10.0m/s2
t=?
dy = 0 m
a
dx = 25.7 m
vf = vi + at
t = (vf – vi) / a
t = (-18.8 – 18.8) / (-10.0)
t = 3.76 s
vy
+
Θ=70o
vx
dx = vixt + ½ at2
dx = (6.84)(t) + ½ (0)t2
dx = (6.84)(3.76) + 0
dx
+