PHYS 1441 – Section 004
Lecture #7
Wednesday, Feb. 11, 2004
Dr. Jaehoon Yu
–
–
–
•
Projectile Motion
Maximum Range and Height
Reference Frame and Relative Velocity
Chapter four: Newton’s Laws of Motion
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Newton’s First Law of Motion
Newton’s Second Law of Motion
Gravitational Force
Newton’s Third Law of Motion
Today’s homework is homework #4, due 1pm, next Wednesday!!
1st term exam in the class at 1pm, Monday, Feb. 23!!
Wednesday, Feb. 11, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
1
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
r  r f  r i
• Average Velocity:
r
r f  ri
v

t
t f  ti
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednesday, Feb. 11, 2004
v  lim
t 0
r
dr

t
dt
How is each of
these quantities
defined in 1-D?
v
v f  vi
a

t
t f  ti
v d v d  d r  d 2 r
a  lim


 2


t 0 t
dt dt  dt  dt
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
Projectile Motion
•
A 2-dim motion of an object under the gravitational acceleration with the
assumptions
– Gravitational cceleration, -g, is constant over the range of the motion
– Air resistance and other effects are negligible
•
A motion under constant acceleration!!!!  Superposition of two motions
– Horizontal motion with constant velocity and
– Vertical motion under constant acceleration
Show that a projectile motion is a parabola!!!
vxi  vi cos 
r
r
r
r
a  ax i  a y j   g j
1
y f  v yi t    g  t 2
2
1 2
 vi sin i t  gt
2
Wednesday, Feb. 11, 2004
v yi  vi sin i
ax=0
Plug in the
t above
In a projectile motion, the only
acceleration is gravitational one whose
direction is always toward the center of
the earth (downward).
x f  vxi t  vi cos i t t 
xf
vi cos i
 xf
 1  xf
  g 
y f  vi sin  i 
 vi cos  i  2  vi cos  i

 2
g

x f
y f  x f tan  i  
2
2

 2vi cos  i 
PHYS 1441-004, Spring 2004 What kind of parabola is this?
Dr. Jaehoon Yu
3



2
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
component in 2-dim,
because the ball is at y=0
position when it completed
it’s flight.
Wednesday, Feb. 11, 2004
80
 t  0 or t 
 8 sec
g
 t  8sec
x f  vxi t  20  8  160  m
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
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!!
vi
Since no
acceleration in
x, it still flies
even if vy=0
v yf  v yi  a y t  vi sin   gt A  0

h
R  vxi 2t A 
 v sin i 
2vi cos i  i

g


t A 
vi sin 
g
y f  h  v yi t 
1
g t2
2
 v sin  i  1  vi sin  i 
  g 

y f  vi sin  i  i
 g  2  g 
 vi 2 sin 2 i 
PHYS 1441-004, Spring 2004
R
Wednesday, Feb. 11, 2004 

g
Dr. Jaehoon Yu


 vi 2 sin 2  i 

y f  

 2g 
5
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


Wednesday, Feb. 11, 2004




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 1441-004, Spring 2004
Dr. Jaehoon Yu
6
Example for Projectile Motion
•
A stone was thrown upward from the top of a building at an angle of 30o to
horizontal with initial speed of 20.0m/s. If the height of the building is 45.0m,
how long is it before the stone hits the ground?
vxi  vi cos   20.0  cos30o  17.3m / s
v yi  vi sin i  20.0  sin 30o  10.0m / s
y f  45.0  v yi t 
t
20.0 
1 2
gt
2
gt 2  20.0t  90.0  9.80t 2  20.0t  90.0  0
 202  4  9.80  (90)
2  9.80
t  2.18s or t  4.22s
•
t  4.22s
Only
physical
answer
What is the speed of the stone just before it hits the ground?
vxf  vxi  vi cos   20.0  cos 30o  17.3m / s
v yf  v yi  gt  vi sin i  gt  10.0  9.80  4.22  31.4m / s
v  vxf 2  v yf 2  17.32   31.4 2  35.9m / s
Wednesday, Feb. 11, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
7