PHYS 1443 – Section 001
Lecture #5
Tuesday, June 5, 2007
Dr. Jaehoon Yu
•
Motion in Two Dimensions
–
–
–
Motion under constant acceleration
Projectile Motion
Maximum ranges and heights
Today’s homework is HW #3, due 7pm, Friday, June 8!!
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1
Announcements
• Quiz results
– Class average: 9.9/14
• Equivalent to: 70.7/100
– Top score: 13/14
• Quiz this Thursday, June 7
– At the beginning of the class
– CH 1 – what we cover tomorrow
• Mail distribution list
– 20 of you are on the list as of last night
– Extra credit
• 3 points if done by tomorrow Wednesday, June 7
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
2
Kinematic Quantities in 1D and 2D
Quantities
1 Dimension
2 Dimension
Displacement
x  x f  xi
r  rf  ri
Average Velocity
x x f  xi
vx 

t
t f  ti
rf  ri
r
v

t
t f  ti
Inst. Velocity
x dx
v x  lim

t  0 t
dt
r
dr
v  lim

t 0 t
dt
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
v f  vi
v
a

t
t f  ti
Inst. Acc.
Tuesday, JuneWhat
5, 2007 is
v dv d 2r
vx dvx d 2 x

 2
a x  lim

 2 a  lim

t

0
t 0 t
t dt dt
dt dt
the differencePHYS
between
and2007
2D quantities?
1443-001,1D
Summer
Dr. Jaehoon Yu
3
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r
r
r
rf  x f i  y f j
r
r
r
ri  xi i  yi j
• Velocity vectors in x-y plane:
r
r
r
vi  vxi i  v yi j
r
r
r
v f  vxf i  v yf j
Velocity vectors in terms of acceleration vector
X-comp
v xf  vxi  axt
Y-comp

v yf  v yi  a y t
 

r
r
r
r
r
r
r
v f   vxi  axt  i   v yi  a yt  j  vxi i  v yi j  ax i  a y j t 
r r
 vi  at
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
4
2-dim Motion Under Constant Acceleration
• How are the position vectors written in acceleration
vectors?
Position vector
components
Putting them
together in a
vector form
Regrouping
the above
Tuesday, June 5, 2007
1 2
x f  xi  vxi t  ax t
2
r
r
r
rf  x f i  y f j 
1 2
y f  yi  v yi t  a y t
2
r 
r
1
1

2
2
  xi  vxit  axt  i   yi  v yit  a yt  j
2
2

 




r 2
r
r
r
r
1 r
 xi i  yi j  vxi i  v yi j t   ax i  a y j  t
2
r r
1r 2
 ri vit  at
2
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
5
Example for 2-D Kinematic Equations
A particle starts at origin when t=0 with an initial velocity v=(20i-15j)m/s.
The particle moves in the xy plane with ax=4.0m/s2. Determine the
components of the velocity vector at any time t.
vxf  vxiaxt  20 4.0t  m / s  v yf  v yi  a y t  15 0t  15  m / s 
Velocity vector
r
r
r
r
r
v  t   vx  t  i v y  t  j   20  4.0t  i 15 j ( m / s )
Compute the velocity and speed of the particle at t=5.0 s.
r
r
r
r
r
r
r
vt 5  vx,t 5i v y ,t 5 j   20  4.0  5.0  i 15 j  40i  15 j m / s

r
speed  v 
Tuesday, June 5, 2007
 vx 
2
  vy 
2

 40
2
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu

  15  43m / s
2
6
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
  tan 
 vx
1

1  15 
1  3 
o

tan

tan


21

 


40
 8 



Determine the x and y components of the particle at t=5.0 s.
1 2
1
x f  vxi t  ax t  20  5   4  52  150( m)
2
2
y f  v yi t  15 5  75 (m)
Can you write down the position vector at t=5.0s?
r
r
r
r
r
r f  x f i  y f j  150i 75 j  m 
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
7
Projectile Motion
• A 2-dim motion of an object under
the gravitational acceleration with
the following assumptions
– Free fall acceleration, g, is constant
over the range
of the motion
r
r
• g  9.8 j  m s 2 
– Air resistance and other effects are
negligible
• A motion under constant
acceleration!!!!  Superposition
of two motions
– Horizontal motion with constant
velocity ( no acceleration )
– Vertical motion under constant
acceleration ( g )
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
8
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos 
v yi  vi sin i
y-component
a axi  a y j   gj
ax=0
x f  vxi t  vi cos i t
t
xf
vi cos i
In a projectile motion,
the only acceleration is
gravitational one whose
direction is always
toward the center of the
earth (downward).
1 2
1
2

v
sin

t

gt
y f  v yi t    g  t
i
i
2
2
Plug t into
the above
xf

xf

 1 
y f  vi sin i 
  2 g  v cos  
i 
 i
 vi cos i 

g
y f  x f tan  i  
2
2
2
v
cos
i
 i
Tuesday, June 5, 2007
 2
x f


PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
2
What kind of parabola is this?
9
Projectile Motion
Tuesday, June 5, 2007
ThePHYS
only
acceleration in this
1443-001, Summer 2007
Yu
motion. Dr.ItJaehoon
is a constant!!
10
Example for Projectile Motion
A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of
flight and the distance the ball is from the original position when landed.
Which component determines the flight time and the distance?
1
y f  40t    g  t 2  0m
2
Flight time is determined
by y component, because
t 80  gt   0
the ball stops moving
when it is on the ground
after the flight.
Distance is determined by x
components in 2-dim,
because the ball is at y=0
position when it completed it’s
flight.
Tuesday, June 5, 2007
So the possible solutions are…
80
 t  0 or t 
 8 sec
g
Why isn’t 0
 t  8sec
the solution?
x f  vxi t  20  8  160  m
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
11
Horizontal Range and Max Height
• Based on what we have learned in the previous pages, one
can analyze a projectile motion in more detail
– Maximum height an object can reach
– Maximum range
What happens at the maximum height?
At the maximum height the object’s vertical
motion stops to turn around!!
v yf  v yi  a y t
vi

h
 vi sin   gt A  0
Solve for tA
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
vi sin 
t A 
g
12
Horizontal Range and Max Height
Since no acceleration is in x direction, it still flies even if vy=0.
 vi sin i 
R  vxi t  vxi  2t A   2vi cos i 

g


 vi 2 sin 2 i 


Range
R

y f  h  v yi t 
Height
g


1
 vi sin i  1  vi sin  i 
2
  g  t  vi sin i 

  g
g
g
2

 2 

 vi 2 sin 2 i 
yf  h  

2g


Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
13
2
Maximum Range and Height
• What are the conditions that give maximum height and
range of a projectile motion?
 vi 2 sin 2  i
h  
2g

 vi 2 sin 2 i 

R  

g


Tuesday, June 5, 2007




This formula tells us that
the maximum height can
be achieved when i=90o!!!
This formula tells us that
the maximum range can
be achieved when
2i=90o, i.e., i=45o!!!
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
14
Example for a Projectile Motion
• A stone was thrown upward from the top of a cliff at an angle of 37o
to horizontal with initial speed of 65.0m/s. If the height of the cliff is
125.0m, how long is it before the stone hits the ground?
vxi  vi cos   65.0  cos 37  51.9m / s
v yi  vi sin i  65.0  sin 37  39.1m / s
1 2
y f  125.0  v yi t  gt
2
Becomes
gt 2  78.2t  250  9.80t 2  78.2t  250  0
t
78.2 
 78.22  4  9.80  (250)
2  9.80
t  2.43s or t  10.4s
t Tuesday,
10.June
4s5, 2007
Since negative time does not exist.
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
15
Example cont’d
• What is the speed of the stone just before it hits the ground?
v xf  v xi  vi cos   65.0  cos 37  51.9m / s
v yf  v yi  gt  vi sin i  gt  39.1  9.80 10.4  62.8m / s
v  vxf  v yf  51.9   62.8  81.5m / s
2
2
2
2
• What are the maximum height and the maximum range of the stone?
Do these yourselves at home for fun!!!
Tuesday, June 5, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
16