PHYS 1441 – Section 002
Lecture #7
Wednesday, Sept. 29, 2010
Dr. Jaehoon Yu
•
•
•
•
Wednesday, Sept. 29,
2010
Projectile Motion
Maximum Range and Height
Force
Newton’s Second Law
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
Announcements
• First term exam results
– Class Average: 52/96
• Equivalent to: 55/100
– Top score: 94/96
• Evaluation policy
–
–
–
–
–
–
Homework: 30%
Final comprehensive exam: 25%
Better of the two non-comprehensive term exam: 20%
Lab: 15%
Quizzes: 10%
Extra credit: 10%
• Colloquium today at 4pm in SH101
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
3
Special Project for Extra Credit
• Show that the trajectory of a projectile motion
is a parabola!!
– 20 points
– Due: Wednesday, Oct. 6
– You MUST show full details of your OWN
computations to obtain any credit
• Beyond what was covered in page 8 of this
lecture note!!
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
4
What is the 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 s2
• ax  0 m s 2 and a y  9.8 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 ) vxf  v x 0
– Vertical motion under constant acceleration ( g )
Wednesday, Sept. 29, 2010


v yf  v y0  PHYS
a y t 1441-002,
 v y0 Fall
 2010
9.8
t
Dr. Jaehoon
5
Projectile Motion
Maximum height
Wednesday, Sept. 29,
2010
The only acceleration in this
motion. It is a constant!!
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
Kinematic Equations for a projectile motion
y-component
x-component
ur
2
a y   g  9.8 m s
ax  0
v

v

gt
v v
y
yo
x
xo
x  vxo t
v v
2
x0
2
xo
x  vxot
Wednesday, Sept. 29,
2010
y 
1
2
v


v
t
yo
y
v  v  2gy
2
y
2
yo
y  v yot  gt
1
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
7
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos
v yi  vi sin  i
y-component
r
r
r
r
a ax i  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
y f  v yit 
g t  vi sin  it  gt
2
2
Plug t into
the above


 1 
xf
xf
y f  vi sin  i 
  2 g  v cos 
 i
 vi cos  i 
i 

 2
g
y f  x f tan i  
xf
2
2

 2vi cos i 
Wednesday, Sept. 29, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
What kind of parabola is this?
8
Example for a Projectile Motion
A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of
flight and the distance from the original position when the ball lands.
Which component determines the flight time and the distance?
Flight time is determined
by the y component,
because the ball stops
moving when it is on the
ground after the flight.


So the possible solutions are…
Distance is determined by the x
component in 2-dim, because
the ball is at y=0 position when it
completed it’s flight.
Wednesday, Sept. 29,
2010
 
y f  40t  1 g t 2  0m
2
t 80  gt  0
80
t  0 or t 
 8sec
g
Why isn’t 0
 t  8sec
the solution?

x f  vxit  20  8  160 m
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
9
Ex.3.9 The Height of a Kickoff
A placekicker kicks a football at an angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
10
First, the initial velocity components
v0  22 m s
v0 y
  40o
v0 x


sin  22 m s sin 40
v0 x  vo cos  22 m s cos 40  17 m s
v0 y  vo
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
o
o
 14 m s
11
Motion in y-direction is of the interest..
y
ay
vy
v0y
?
-9.8 m/s2
0 m/s
+14 m/s
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
t
12
Now the nitty, gritty calculations…
y
ay
vy
v0y
?
-9.80 m/s2
0
14 m/s
t
What happens at the maximum height?
The ball’s velocity in y-direction becomes 0!!
And the ball’s velocity in x-direction? Stays the same!! Why?
Because there is no
acceleration in xdirection!!
Which kinematic formula would you like to use?
v  v  2a y y
2
y
2
oy
y
Wednesday, Sept. 29,
2010
Solve for y
0  14 m s

2 9.8m s
2


y
v 2y  voy2
2

2a y
 10 m
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
13
Ex.3.9 extended: The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
14
What is y when it reached the max range?
y
ay
0m
-9.80 m/s2
Wednesday, Sept. 29,
2010
vy
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
voy
t
14 m/s
?
15
Now solve the kinematic equations in y direction!!
y
ay
0
-9.80 m/s2
y  voy t  a y t
1
2
Two soultions
2
t 0
voy  12 a y t  0
Wednesday, Sept. 29,
2010
Since y=0
vy
voy
t
14 m/s
?

0  voy t  a y t  t voy  12 a y t
2
1
2

or
Solve
for t
t
voy
1
ay
2

2voy
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
ay
2 14
 2.9s

9.8
16
Ex.3.9 The Range of a Kickoff
Calculate the range R of the projectile.
x  voxt  axt  vox t  17 m s 2.9 s  49 m
1
2
Wednesday, Sept. 29,
2010
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
17
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  cos37o  51.9m / s
v yi  vi sin  i  65.0  sin37o  39.1m / s
y f  125.0  v t  1 gt 2
yi
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 Wednesday,
10.4s
Sept. 29,
2010
Since negative time does not exist.
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
18
Example cont’d
• What is the speed of the stone just before it hits the ground?
vxf  vxi  vi cos  65.0  cos37o  51.9m / s
v yf  v yi  gt  vi sin  i  gt  39.1 9.80 10.4  62.8m / s
v 
v v 
2
xf
2
yf

  81.5m / s
51.9  62.8
2
2
• What are the maximum height and the maximum range of the stone?
Do these yourselves at home for fun!!!
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
19
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 v0 y  a y t  v0 sin  0  gt A  0
Solve for tA
Wednesday, Sept. 29,
2010
v0 sin  0
t A 
g
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Time to reach to the
maximum height!!
20
Horizontal Range and Max Height
Since no acceleration is in x direction, it still flies even if vy=0.
 v0 sin  0 
R  v0 xt v0 x 2t A   2v0 cos  0 

g


 v02 sin 20 
R

g


Range
 v sin 0  1  v0 sin  0 
y f  h  v t  1 g t 2  v0 sin  0  0
 g


0y
g
g

 2 
2
 
Height
 v02 sin 2  0 
yf  h  

2g


Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
21
2
Maximum Range and Height
• What are the conditions that give maximum height and
range of a projectile motion?
 v02 sin 2 0 
h

2g


 v02 sin20 
R

g


Wednesday, Sept. 29,
2010
This formula tells us that
the maximum height can
be achieved when
θ0=90o!!!
This formula tells us that
the maximum range can be
achieved when 2θ0=90o,
i.e., θ0=45o!!!
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
22