Physics I
Class 02
Two-Dimensional Motion
Rev. 04-Jan-04 GB
02-1
One-Dimensional Motion with
Constant Acceleration - Review
Basic Equations
1. v  v 0  a t  t 0 
2. x  x 0  v 0 ( t  t 0 )  2 a ( t  t 0 )
1
2
Derived Equations
3. x  x 0  2 ( v 0  v)( t  t 0 )
1
4. x  x 0  v( t  t 0 )  2 a ( t  t 0 )
1
2
2
2
v

v
5.
0  2a  x  x 0 
02-2
Simple Example of 2D Motion
Throw a ball straight up (1D).
Throw a ball across the room (2D).
02-3
Motion in Two Dimensions
 Motion in the X (horizontal) direction is
independent of motion in the Y (vertical) direction.
(Neglecting air resistance.)
 Strategy: Break the problem into two parts, one for
the X motion and one for the Y motion.
 Handle each part like one-dimensional motion.
02-4
Two-Dimensional Equations
Basic X Equations
1X. v x  v 0 , x  a x t  t 0 
2X. x  x 0  v 0 , x ( t  t 0 )  2 a x ( t  t 0 )
1
2
Basic Y Equations
1Y. v y  v 0 , y  a y t  t 0 
2Y. y  y 0  v 0 , y ( t  t 0 )  2 a y ( t  t 0 )
1
2
02-5
A Special Case of 2D Motion Projectile Motion
Y
X
By usual convention, we choose X to the right and Y
up when we study projectile motion.
Throwing a baseball and shooting a cannon are examples of
projectile motion. We neglect air resistance and assume that the
only force is gravity (down).
The acceleration in the X direction is zero: a x  0
The acceleration in the Y direction is constant and directed down:
a y   9 .8 m / s 2
02-6
Projectile Motion Vx and Vy
02-7
Problem Solving Strategy for
Projectile Motion
Make a table, see what you know and what you need to find.
a
v0
x0 or y0
vf
xf or yf
t-t0
X
Y
SAME
SAME
The common factor in both the X and Y equations is the
time at which something happens (last row).
02-8
Resolving Vectors into
Components
o vo, y
sin   
h
vo
a vo, x
cos  
h
vo
so
so
vo, y  vo sin 
vo, x  vo cos

V0
V0,Y

V0,X
Would Vo,y still be related to sine if we were given the
other angle, ?
02-9
Example of Projectile Motion Hit the Falling Target
02-10
Hit the Falling Target Diagram
The target will drop at the instant
the ball leaves the launcher.
h

d
The objective is to adjust the angle  so that the ball
hits the falling target.
02-11
Hit the Falling Target
Table of Kinematic Quantities
a
v0
x0 or y0
vf
xf or yf
t-t0
X ball
0
v0 cos()
0
v0 cos()
d
?
Y ball
-g
v0 sin()
0
?
?
SAME
Y target
-g
0
h
?
Same as ball.
SAME
We have all data in the “X ball” column except time.
Solve for that first.
02-12
Hit the Falling Target
Solving for Time
d
x f  x 0  v 0 cos( ) ( t  t 0 )  ( t  t 0 ) 
v 0 cos( )
We don’t care what vf is for the ball or target (Y).
a
v0
x0 or y0
vf
xf or yf
t-t0
X ball
0
v0 cos()
0
v0 cos()
d
d/[v0 cos()]
Y ball
-g
v0 sin()
0
(don’t care)
?
SAME
Y target
-g
0
h
(don’t care)
Same as ball.
SAME
Next, use the kinematics equation to find yf.
02-13
Hit the Falling Target
Solving for Final Y Position
Y ball:
Y target:

d
1 
d
y f  0  v 0 sin( )
 g
v 0 cos( ) 2  v 0 cos( ) 

1 
d
yf  h  0  g 
2  v 0 cos( ) 
2
2
Setting the two expressions equal, the “g” terms cancel and we are left with
d
h
sin( )
 h OR tan( ) 
cos( )
d
02-14
Hit the Falling Target
Final Solution for Launch Angle
h

d
h
tan( )  means that we aim directly at the target.
d
Are we “ignoring” gravity? Where did v0 go?
02-15
Class #2
Take-Away Concepts
1.
2.
3.
4.
X and Y motions are independent.
In projectile motion problems, the acceleration is
constant = 9.8 m/s2 down (normally -Y direction).
Strategy for solving projectile motion problems:
Create a table, fill in known quantities, work on finding
unknown quantities.
Use time to connect information from one column to
another.
02-16
Class #2
Problems of the Day
_______1. Two cannon crews, A and B, are practicing shooting their
cannons in a large, flat field. Crew A points their cannon at an
angle of 30° up from horizontal. Crew B points their cannon 60°
up from horizontal. They both fire at the same time and they
notice that their respective cannon balls hit the ground at the same
time. Neglecting air resistance, and assuming both cannon balls
start from the same height when fired, which cannon ball reaches
the greatest maximum height?
(Do not assume that the cannons and cannon balls are identical.)
A) The cannon ball from crew A goes higher.
B) The cannon ball from crew B goes higher.
C) Both cannon balls reach the same maximum height.
02-17
Answer to Problem 1 for Class #2
The answer is C.
Consider only the vertical motion of the cannon balls. Both cannon
balls start up at the same time and both hit the ground at the same
time. They are both subject to the same acceleration due to gravity.
The only way they could hit at the same time would be for both
cannon balls to reach the same height.
The horizontal motions of the two cannon balls will be different, but
the horizontal motions are independent of the vertical motions.
02-18
Class #2
Problems of the Day
2. Crew A from Problem 1 aims their cannon at an angle of 30° up from
horizontal and fires it as shown below in a large, level field. The initial
speed of the cannon ball is 196 m/s. Neglect air resistance and assume that
the cannon ball starts at ground level for simplicity. Use g = 9.8 m/s 2.
What is the maximum height reached by the cannon ball?
30°
02-19
Answer to Problem 2 for Class #2
The first step is to determine the vertical component of the initial velocity. This is the
initial speed times sin of the angle from horizontal. V0y = (196) sin(30) = 98 m/s.
The second step is to realize that the vertical component of velocity will be zero at the
highest point.
What variables do we know? y0 = 0, v0y = 98, t0 = 0 (start the clock when the cannon
is fired), vy = 0 and ay = –9.8 . We don’t know y (what the problem asks for) or t.
Method A: Use eq. 1 to solve for t: 0 = 98 – 9.8 * t . t = 10.0 . Then use eq. 2 to
solve for y: y = 0 + 98 * 10 – 0.5 * 9.8 * 102 = 490 m.
Method B: We realize that we know everything in eq. 5 except y, which is what we
want. (0)2 = (98)2 + 2.0 * (–9.8) * y. y = (98)2 /(2.0 * 9.8) = 490 m.
02-20
Activity #2
Projectile Motion
Objectives of the Activity
1. Making sure VideoPoint is installed and working
correctly on your laptop.
2. Learning how to use VideoPoint.
3. Using VideoPoint to study projectile motion.
02-21
Class #2 Optional Material
Galileo
Galileo Galilei (1564-1642)
Galileo studied the motion of freefalling bodies and bodies falling along
inclined planes. He was the first to
determine that the distance traversed by
a falling body in equal time intervals
follows the series 1, 3, 5, 7, 9, …
(This is equivalent to saying that the
total displacement is proportional to the
total time squared.)
02-22
Aristotle’s Error
Aristotle
384-322 B.C.E.
Aristotle was unquestionably a genius, but he missed the connection
between theory and observation. He taught that a projectile travels in
a straight line until it loses the motion imparted to it, then drops
straight down. Medieval scholars invented the term “impetus” for the
imparted motion. A simple observation of a person throwing a rock
disproves this theory, but Aristotle did not think to do that.
02-23
Parabolic Trajectory
Galileo deduced from his observations that horizontal and vertical motions
are independent. From that he deduced that projectiles travel in curved
paths and that these curves must be parabolas. He published his results in
Discourses on Two New Sciences, 1638.
Galileo is also responsible for an early form of the Principle of Relativity,
which was not revised until Einstein.
02-24