Motion in Two Dimensions

 Intro to projectile motion

 Vector: a mathematical object. Vector fields.
Components of vectors
 ABC of projectile motion.
 More on motion in 2D
 More on vectors. Adding vectors.
 Circular motion. Tangential and centripetal
acceleration

February 5-8, 2013

Hippie-Jump Any 2 D motion can be decomposed
Into two independent motions:
One along the horizontal and one
Along the vertical/

→ we break down a 2D problem

To 2 1D

→ The x-motion takes place as if the

Y-motion isn’t happening. The
Y motion takes place independent
Of whatever is happening in
The x direction
→ We get 2 sets of
Kinematics equation

What are theses equations

For the jumper ?
Video
What are the 2 sets of equations here?
One set long the x-axis and one along the y-axis.
Say the speed is 10 m/s (to simplify. Its actually about 5m/s)
How long to reach the ground? Distance covered?
 Y
parabola

Voy
Vi
Vox
X
●
Along vertical (Y-axis) ●
Along horizontal no acceleration
●
free-fall
a= 0
Acceleration is -10m/s/s Δx is the horizontal displacement (x-axis)
Δy is the vertical displacement ( y-axis ) We suppose initial position is 0.
We suppose initial position is 0. Vix is the initial velocity along x
Viy is the initial velocity along y (Vix is the x-component of Vi the initial velocity)
(Viy is the y-component of Vi the initial velocity)
ΔY =Viy t - 5t2 Δx= Vix t
Vy=Viy -10 t Vx = Vix
Vy 2 =Viy 2 -20ΔY Velocity stays constant.

Time t is the same !

 BIG IDEA
A two-dimensional problem
can be considered as 2
one-dimensional problems.
A body motion has 2 components
completely independent of each other.
2 components means : we will use trig to find the components of the velocities along the path.
A horizontal component (cosine if you the standard notation) and vertical component
(sine if you use the standard notation).
http://www.phy.hk/wiki/j/Eng/projectile/
projectile_js.htm
Example A simple projectile motion: initial velocity is horizontal
Voy= 0 and Vox=115 m/s. The vertical displacement is -1050m.
Determine the time required for the care package to hit the ground. (use the y-axis free-fall equations)
The final vertical velocity Vy (free-fall equation)
The horizontal displacement (equation for motion at constant speed along X-axis)
Optional: Using the components of the final velocity Vox and Vy find the
final velocity V (direction and magnitude).
Will they hit at the same time ?
Will the final speed be the same?
Both packages are dropped at the same time.
 Video from virginia tech
free_fall_projectile

See exploration software

Racing balls # 6

Video
Shoot n drop

1 motion but 2 independent components.

 Source: Paul Hewitt (conceptual physics). .d = 10m and y=1m what is the horizontal velocity to clear the net?
Source: Paul Hewitt (conceptual physics). The Earth is curving away : 8000m run and 5 m fall.
What is the minimum horizontal velocity forn a launched from mountain so it does not crash (sic Newton)
 X

https://phet.colorado.edu/sims/html/vector-addition/latest/vector-addition_en.html

February 5-8, 2013

WHICH quantity is not a vectorWHICH quantity is a scalar

1. force 1. distance

2. speed 2. velocity

3. acceleration 3. acceleration

4. displacement 4. displacement

16
 This is a
Vector field

Example:
Gravitational
Vector field

February 5-8, 2013

https://ophysics.com/k3.html
Video vector field
→ think fluid

Examples of vector field. Vortex, gravity, electric field

Review:
https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-trig-ratios-intro/v/basic-trigonometry

adj SOH CAH TOA

hyp opp hyp

opp adj

Find A

There are
Tutorials in the
dropbox

→ components
Of vectors

19
To find components: start from the tail and get to the head along
The x-axis then the y-axis. Or along y-axis the the x-axis.

February 5-8, 2013

Show simulation html → vector components, multiplication

Textbook 1.5 and 1.6

Study vectors. unit1

February 5-8, 2013

https://ophysics.com/k3.html
February 5-8, 2013
See SIMULATION Textbook 1.6

 Ay 
A 2
A A
x
2
y and   tan  1

 Ax 

Or,

February 5-8, 2013

 Bring polar graph in class
Standard notation →
stand
The angle is between 0 and 360.
If you use the standard notation:
x-component = cosine x magnitude
y-component = sine x magnitude
The initial velocity has 2 components
Forward (horizontal) and vertical.
The vertical component decreases to 0
At apogee and the horizontal component
Stays constant.
Source:
http://sdsu-physics.org/physics180/physics180A/units/unit1/
chapter3.html
 Θ

Along the horizontal (x-axis) the x- component of the

Velocity stays the same. Vix = Vx

Along the vertical (y-axis) the y-component of the velocity Changes at a constant
rate : Vy= Viy – 10t so decreases at a rate of 10m/s/s
Vix = Vo cos(Θ)
●
Viy= Vo sin(Θ)
https://www.physicsclassroom.com/class/vectors/Lesson-2/Horizontal-and-Vertical-Components-
of-Velocity
 Y
parabola

Viy
Vi https://ophysics.com/k9.html
Vix http://www.phy.hk/wiki/j/Eng/projectile/
projectile_js.htm
X
●
Along vertical (Y-axis) ●
Along horizontal no acceleration
●
free-fall
a= 0
Acceleration is -10m/s/s Δx is the horizontal displacement (x-axis)
Δy is the vertical displacement ( y-axis ) We suppose initial position is 0.
We suppose initial position is 0. Vix is the initial velocity along x
Viy is the initial velocity along y (Vix is the x-component of Vi the initial velocity)
(Viy is the y-component of Vi the initial velocity)
ΔY =Viy t - 5t2 Δx= Vix t
Vy=Viy -10 t Vx = Vix
Vy 2 =Viy 2 -20ΔY Velocity stays constant.

Time t is the same !

a y Viy Vy t
-10
Remember that acceleration is force per unit mass.
As long as there is a force acting -. acceleration
The acceleration due to gravity is the weight per unit mass.

A ball is launched with an initial velocity v0 as shown. Which one of
the following arrows best represents the direction of the
acceleration at point A? B ? C?

a)

b)

c)

d)

e) The acceleration at point A is zero m/s2.

3.3 Projectile Motion

Example 6 The Height of a Kickoff

A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is Vo= 22 m/s. Ignoring air resistance
1) Along x: Vix=? and Viy=?
2) Use the free-fall equations to find the maximum height H
3) the time of flight
4) Use the X-axis equation (constant velocity) to find the range
5) consider the time when the ball hits the ground (use the time of flight)
Find the components of the final velocity Vx and Vy at that time.
6) Find the vector final velocity V given Vx and Vy. (magnitude and direction)

February 5-8, 2013

Maximum height H

y ay vy voy t

? -9.80 m/s2 0 14 m/s

v v
2
y
2
oy
y
v  v  2a y y
2
y
2
oy 2a y
0  14 m s 
2
y  10 m

2  9 .8 m s 2

February 5-8, 2013
Help Peter

February 5-8, 2013

 Here are the physical quantities that describe
Projectile motion:
- x is the horizontal position, y the vertical position
- R is the range of the projectile
- H is the maximum height reached by projectile
- Vix and Viy the components of the initial velocity Vo
- Vx and Vy the components of the velocity along the path
- t is the time elapsed

Textbook3.6
 See textbook
Whats the angle for
max height
And range?

What is R and h ?
Trajectory of Projectile Motion
 Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0y = v0 sinθ0
 Horizontal motion:
x
x=0+v 0 x t ⇒ t=
v0 x
 Vertical motion:
1
y=0+v 0 y t− 2 gt 2

( ) ( )
2
x g x
y=v 0 y −
v0 x 2 v0 x
g 2
y= x tan θ0 − x
2 v 2 cos 2 θ 0
0

 Parabola;
 θ0 = 0 and θ0 = 90 ?

February 5-8, 2013

Good runners → good long jumper
https://commonsciencespace.com/running-out-of-sugar-hitting-
wall/
See projectile_motion_Summer200.pdf
To explain

Hunter wants to shoot the coconut.

Should he aim above? Below? At?

See exploration software # 7 and video monkey @ MIT

Textbook problems
 Motion in two dimensions
 Motions in each dimension are independent components
 Constant acceleration equations
1
⃗v =⃗v 0 +⃗a t ⃗r −⃗r =⃗v 0 t+ 2 ⃗a t 2
 Constant acceleration equations hold in each dimension
v x =v 0 x +a x t v y=v 0 y +a y t
1 1
x−x 0 =v 0 x t+ 2 a x t 2 y− y 0 =v 0 y t+ 2 a y t 2
v 2 =v 2 +2 a x ( x−x 0 ) v 2 =v
x 0x 2 +2 a y ( y − y0 )
y 0y
 t = 0 beginning of the process;
 ⃗a =a x ^i +a y ^j where ax and ay are constant;
 Initial velocity ⃗v 0 =v 0 x ^i +v 0 y ^j initial displacement ⃗r 0 =x 0 ^i + y 0 ^j ;
February 5-8, 2013
 Motion of a Turtle
The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in
the horizontal direction does not affect motion in the vertical direction, and vice versa. Imagine a shadow
(projection) along the x-axis and y-axis. They move
At their own velocity Vx and Vy and the positions x=Vx t and y = Vy t

A turtle starts at the origin and moves with the speed of v0=10 cm/s in
the direction of 25° to the horizontal.
(a) Find the coordinates of a turtle 10 seconds later.
(b) How far did the turtle walk in 10 seconds?

February 5-8, 2013

 Motion of a Turtle
Notice, you can solve the
equations independently for the
horizontal (x) and vertical (y)
components of motion and then
combine them!
⃗v 0 =⃗v x +⃗v y
 X components:
v0 x  v0 cos 25   9.06 cm/s  x  v0 x t  90.6 cm
 Y components:
v0 y  v0 sin 25  4.23 cm/s  y  v0 y t  42.3 cm
 Distance from the origin:
d=√ Δx2 +Δy 2=100.0 cm
February 5-8, 2013
 Lets go back to vectors.
Finding the components:
→ How much of the vector is along X
→ How much of the vector is along Y

D B

If you cut and paste

The vector in the polar graph,
C A X-component is cosine
Y-component is the sine.
 ?

Ref: https://www.youtube.com/watch?
v=370j60wSwlo
 Position and Displacement
 In one dimension
Δx=x 2 (t 2 )−x 1 (t 1 )
x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 m
Δx = +1.0 m + 3.0 m = +4.0 m Δ ⃗r =⃗r 2−⃗r 1

 In two dimensions
 Position: the position of an object is

described by its position vector ⃗r (t )

--always points to particle from origin.
 Displacement: Δ ⃗r =⃗r 2−⃗r 1
Δ ⃗r =( x 2 ^i + y 2 ^j )−( x 1 ^i + y 1 ^j )
=¿( x −x ) ^i+( y − y ) ^j
2 1 2 1
=¿ Δx ^i +Δy ^j
February 5-8, 2013
 Average & Instantaneous Velocity
 Average velocity Δ ⃗r
⃗v avg ≡
Δt
Δx ^ Δy ^
⃗v avg = i+ j=v avg , x ^i +v avg, y ^j
Δt Δt
 Instantaneous velocity

Δ ⃗r d ⃗r
⃗v ≡lim ⃗v avg =lim =
t →0 t→ 0 Δt dt
d ⃗r dx ^ dy ^
⃗v = = i + j=v x ^i +v y ^j
dt dt dt
 v is tangent to the path in x-y graph;

Textbook3.1
February 5-8, 2013
Average & Instantaneous Acceleration
 Average acceleration Δ ⃗v
⃗a avg≡
Δt
Δv x Δv y
⃗a avg = ^i+ ^j=a ^ ^
avg, x i+aavg , y j
Δt Δt

 Instantaneous acceleration
Δ ⃗v d ⃗v d ⃗v dv x ^ dv y ^
⃗a ≡lim ⃗a avg =lim = ⃗a = = i+ j=a x ^i +a y ^j
t →0 t→ 0 Δt dt dt dt dt

 The magnitude of the velocity (the speed) can change

 The direction of the velocity can change, even though the
magnitude is constant
 Both the magnitude and the direction can change
February 5-8, 2013
Textbook3.2
 Summary in two dimension
 Position ⃗r (t )=x ^i + y ^j
Δ ⃗r Δx ^ Δy ^ ^ ^
 Average velocity avg Δt = Δt i + Δt j=v avg , x i +v avg , y j
⃗
v =

dx dy
 Instantaneous velocity v x≡
v y ≡
dt dt
Δ⃗r d ⃗r dx ^ dy ^
⃗v (t )=lim = = i + j=v x ^i +v y ^j
t →0 Δt dt dt dt
dv x d 2 x dv y d 2 y
ax≡ = 2 a y≡ = 2
 Acceleration dt dt dt dt
Δ⃗v d ⃗v dv x ^ dv y ^
⃗a (t )=lim = = i+ j=a x ^i +a y ^j
t →0 Δt dt dt dt
 ⃗r (t), {⃗v (t), and {⃗a ¿(t)¿ are not necessarily same direction.

February 5-8, 2013

Note the tail to head method. Start from tail → head

Relative to Gnd
The velocity of
The plane has
2 components.

North and west.

The motion has

2 components
Independent from
Each other
If the river is 1,000 wide. How long to reach the other side
https://
ophysics.com/
k11.html

February 5-8, 2013

The vector velocity of the ball has 2 components (perpendicular)
The motion of the ball is defined by the final velocity
=> 1 component is the velocity of the footballer
And 1 component is the velocity of the thrown football.
Note: there are 2 right angle triangles. (Pythagorean and trig)
 How fast the plane is going up ? (vertical component of the velocity)
How fast is plane is moving forward?
(how long it will take for the plane to be over your head)

What are the

components of the
force?

How hard the wind is pushing forward?

How hard the wind is pushing side way? See the picture of the
Sprinter.
February 5-8, 2013
The velocity of the boat has 2 components.
The 2 components are independent from each other.
It is moving @ North and @ East at the same time. independently
The speed of a ferry has a speed of 30mph / river
The speed of the river is 10 mph/ ground

The river is 2 miles wide. How long it takes to cross it.

https://ophysics.com/k11.html
N

February 5-8, 2013

E
 Adding vectors
When finding relative velocity → you are Adding vectors
VB/S = VB/W +VW/S

Vbw is the velocity of the boat

relative to river.
Vws is the velocity of the river
relative to shore
Vbs is the velocity of the boat
relative to gnd → it can crosses!
Adding vectors
The vectors don’t have to be perpendicular.
How to add vectors?

Tail to head method or

Parallelogram method.
February 5-8, 2013
 WAY #1

WAY #2 Attach vectors by their tails. WAY #2

Connect the subtrahend to minuend
 Subtracting vectors → find the displacement
or change in velocity
Attach vectors by tail.

Here an object moves from P1 to P2

.r2 – r1 is the vector displacement.
If you divide by time = average velocity

V2-V1 is the change in velocity

If you divide by the time, you get average acceleration
 P2
P2

P1

What is P2 – P1? ( Attach the vectors tail to tail then

Connect the head of P1 to head of P2)
The change in momentum is proportional The vertical momentum →
To the force needed to change the momentum Huge force over small time
From horizontal to vertical . The force is from the wall Force is friction and propels
It has a horizontal component (N) upward
and vertical component f(static friction)
 Time for some Calculus !
→ go over textbook problems.

Show the displacement vector. This is r2-r1.

Show how to find it [ (x2-x1) , (y2-y1) ]
The average velocity is (r2-r1)/time.
Show the instantaneous velocity.
The acceleration has 2 components.
One changes the direction
And one changes the speed

https://physicscatalyst.com/mech/two-dimensional-motion.php
 Uniform circular motion

Constant speed, or, Motion along a circle:

constant magnitude of velocity Changing direction of velocity

February 5-8, 2013

You spin a ball attached to a string. Which what will it go if you let
Go ? How this related to inertia ? And what if the Sun was to
Disappear , which way Will the Earth go ?

See exploration of
Physical for applet.
The sling.

See
See 2 video circular motion pic
You need to apply a centripetal force
To keep the object on the track
(demo: glass of water and the broom)
Source: the physics of every day phenomena / Mc Graw Hill
February 5-8, 2013
February 5-8, 2013
Max acceleration we can support
Is 10gs.

Space shuttle is 3gs → 3 times

Your Weight crushing you.
 See applets from exploration of physical

DEFINITION OF UNIFORM CIRCULAR MOTION

Uniform circular motion is the motion of an object

traveling at a constant speed on a circular path.
The velocity is a vector.

The speed is constant

But not the velocity !!!
Can you see why ?

The velocity changes in direction.

You need a force to keep the
Plane on the track.
So there is an acceleration !
Force = acceleration !
For an object in uniform circular motion, which one of the
following statements is false ?

a) The velocity of the object is constant.

b) The magnitude of the acceleration of the object is constant.

c) The acceleration is directed radially inward.

d) The magnitude of the velocity is constant.

e) The velocity is directed in a direction that is tangent to the

circular path.
When using the term “uniform circular motion,” what do
we mean by the term “uniform?”

a) The direction of the object’s velocity is constant.

b) The net force on the moving object is zero newtons.

c) The forces acting on the object are uniformly applied

from all directions.

d) The motion occurs without the influence of the

gravitational force.

e) The motion of the object is at a constant speed.

Let T be the time it takes for the object to travel once around the circle
T is the period (s). The distance around a circle= 2 pi r
(or circumference – 360 degrees)
Note: The distance around half a circle = pi r (180 degrees)
The distance around ¼ a circle = 2 pi r /4 (90 degrees)
V is a velocity and its direction changes
Its magnitude is constant and is noted v
Show how the vector V changes
With time. (attach the tails
To origin)

Speed = distance / time

r
Speed = 2 pi R / time
An airplane flying at 115 m/s due east makes a gradual
turn while maintaining its speed and follows a circular
path to fly south. The turn takes 15 seconds to complete.
What is the radius of the circular path?

a) 410 m

b) 830 m
Round to the nearest hundred.

c) 1100 m From east to south that's ¼ of a circular path

(circumference of a circle is 2 pi R)
d) 1600 m

e) 2200 m
5.1 Uniform Circular
Motion

Example

The wheel of a car has a radius of 0.29m and it being rotated

at 830 revolutions per minute on a tire-balancing machine.
Determine the speed at which the outer edge of the wheel is
moving.
REMEMBER ACCELERATION = RATE of CHANGE of velocity

VELOCITY is a vector. VELOCITY CHANGES IF :

- magnitude changes (speed changes) or if

- direction changes !!!

Or BOTH

NOTE: ACCELERATION IS PUSH/PULL per unit mass.

ACCELERATION and FORCE have the same DIRECTION !!!

Acceleration is force per unit mass.

 Circular Motion: Observations
 Object moving along a
curved path with constant
speed
 Magnitude of velocity: same
 Direction of velocity: changing
 Velocity: changing
 Acceleration is NOT zero!
 Net force acting on the
object is NOT zero
 “Centripetal force”
⃗F =m ⃗a
net
February 5-8, 2013
Centripetal acceleration – acceleration is a vector =Δv/Δt

vi
Δv = vf - vi
vf
vi y B
A vf
Δr R
ri rf

O
x

Direction: Centripetal
February 5-8, 2013
 We have similar triangles
Δv Δr vΔr
= so, Δv=
v r r
2
Δv Δr v v
= =
Δt Δt r r
Δv v 2
ar= =
Δt r
 This is the magnitude of the
acceleration. Direction is
centripetal
February 5-8, 2013
Another elegant way to show a = V2/R
Ref: https://www.youtube.com/watch?v=XZaJLBOJPNY&t=2s

If V is x 2 then ΔV is x 2

If V is x 2 then Δt is divided by 2

So if V is x 2 then a =ΔV/Δt is x 4

If V is stays the same but R is x 2

ΔV stays the same

But Δt is x 2 so a =ΔV/Δt is divided by 2

Conclusion : a = V2/R

September 18, 2024

 Uniform Circular Motion
 Velocity:
 Magnitude: constant v
⃗a c ⊥⃗v
 The direction of the velocity is
v2
tangent to the circle a c=
r
 Acceleration: v2
 Magnitude: a c=
r
 directed toward the center of the
circle of motion
 Period:
 time interval required for one
complete revolution of the 2 πr
particle T=
v
February 5-8, 2013
A satellite orbits the Earth in uniform circular motion. What is the direction of
centripetal acceleration of the satellite?

a) Because the centripetal acceleration is a scalar quantity, it doesn’t have a

direction.

b) The centripetal acceleration vector points radially outward from the Earth.

c) The centripetal acceleration vector points radially inward toward the Earth.

d) The centripetal acceleration vector points in the direction of the satellite’s

velocity.

e) The centripetal acceleration vector points in the direction opposite that of the
satellite’s velocity.
 5.2 Centripetal
Acceleration
Example : The Effect of Radius on Centripetal Acceleration

The bobsled track contains turns

with radii of 33 m and 24 m.
Find the centripetal acceleration
at each turn for a speed of
34 m/s. Express answers as
multiples ofg  9.8 m s 2 .

Textbook3.12
A satellite orbits the Earth in uniform circular motion. What is the direction of
centripetal acceleration of the satellite?

a) Because the centripetal acceleration is a scalar quantity, it doesn’t have a directio

b) The centripetal acceleration vector points radially outward from the Earth.

c) The centripetal acceleration vector points radially inward toward the Earth.

d) The centripetal acceleration vector points in the direction of the satellite’s veloci

e) The centripetal acceleration vector points in the direction opposite that of the
satellite’s velocity.
 v2 2 r
ac  v 
r T
4 r
2
ac  2
T
 In an non inertial
frame of reference, the
centripetal force is
The support force
From floor (normal
force)
HAL is watching you !!

The space station was 150 about in radius (500ft)

How fast it was rotating (V) to simulate gravity?
(g=10m/s/s). T?
A truck is traveling with a constant speed of 15 m/s. When the
truck follows a curve in the road, its centripetal acceleration is
4.0 m/s2. What is the radius of the curve?

a) 3.8 m

b) 14 m

c) 56 m

d) 120 m

e) 210 m
While we are in this classroom, the Earth is orbiting the Sun
in an orbit that is nearly circular with an average radius
of 1.50 × 1011 m. Assuming that the Earth is in uniform
circular motion, what is the centripetal acceleration of
the Earth in its orbit around the Sun?

a) 5.9 × 103 m/s2

4 r 2
ac  2
b) 1.9 × 105 m/s2 T

c) 3.2 × 107 m/s2

d) 7.0 × 102 m/s2

e) 9.8 m/s2
A bicycle racer is traveling at constant speed v around a
circular track. The centripetal acceleration of the
bicycle is ac. What happens to the centripetal
acceleration of the bicycle if the speed is doubled to
2v?

a) The centripetal acceleration increases to 4ac.

b) The centripetal acceleration decreases to 0.25 ac.

c) The centripetal acceleration increases to 2ac.

d) The centripetal acceleration decreases to 0.5ac.

e) The centripetal acceleration does not change.

Velocity is a vector It has a magnitude and a direction.
The magnitude can change and not the direction (free-fall)
, the direction can change (uniform circular motion)
and not the magnitude , or both !
5.7.3. A steel ball is tied to the end of a string and swung in
a vertical circle at constant speed. Complete the
following statement: The direction of the instantaneous
velocity of the ball is always

a) perpendicular to the circle.

b) toward the center of the circle.

c) tangent to the circle.

d) radially outward from the circle.

e) vertically downward. © Veronique Lankar 201

February 5-8, 2013
 Summary
 Position ⃗r (t )=x ^i + y ^j
Δ ⃗r Δx ^ Δy ^ ^ ^
 Average velocity avg Δt = Δt i + Δt j=v avg , x i +v avg , y j
⃗
v =

dx dy
 Instantaneous velocity v x≡
v y ≡
dt dt
Δ⃗r d ⃗r dx ^ dy ^
⃗v (t )=lim = = i + j=v x ^i +v y ^j
t →0 Δt dt dt dt
dv x d 2 x dv y d 2 y
ax≡ = 2 a y≡ = 2
 Acceleration dt dt dt dt
Δ⃗v d ⃗v dv x ^ dv y ^
⃗a (t )=lim = = i+ j=a x ^i +a y ^j
t →0 Δt dt dt dt
 ⃗r (t), {⃗v (t), and {⃗a ¿(t)¿ are not necessarily in the same direction.

February 5-8, 2013

 Summary
 If a particle moves with constant acceleration a, motion
equations are 1 2
⃗r f =⃗r i +⃗v i t+ 2 ⃗a t
⃗r f =x f ^i + y f ^j=( x i +v xi t + 2 a xi t 2 ) ^i +( y i + v yi t + 2 a yi t 2 ) ^j
1 1

⃗v =⃗v i +⃗a t
⃗v f (t )=v fx ^i +v fy ^j=(v ix +a x t ) ^i +(v iy +a y t ) ^j
 Projectile motion is one type of 2-D motion under constant
acceleration, where ax = 0, ay = -g.

February 5-8, 2013