PHYS 1441 – Section 002
Lecture #7
Wednesday, Feb. 6, 2013
Dr. Jaehoon Yu
•
•
•
What is the Projectile Motion?
How do we solve projectile motion
problems?
Maximum Range and Height
Today’s homework is homework #5, due 11pm, Tuesday, Feb. 12!!
• 1st term exam
Announcements
– In class, Wednesday, Feb. 13
– Coverage: CH1.1 – what we finish Monday, Feb. 11, plus Appendix
A1 – A8
– Mixture of free response problems and multiple choice problems
– Please do not miss the exam! You will get an F if you miss any
exams!
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
2
Special Project #2 for Extra Credit
• Show that the trajectory of a projectile motion
is a parabola!!
– 20 points
– Due: Monday, Feb. 18
– You MUST show full details of your OWN
computations to obtain any credit
• Beyond what was covered in page 7 of this
lecture note!!
Wednesday, Sept. 29,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
3
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
(
)
•
m s2
• a = 0 m s2 and a y = -9.8 m s2
x
– Air resistance and other effects are
negligible
• A motion under constant
acceleration!!!!  Superposition of
two motions
– Horizontal motion with constant velocity ( no
acceleration ) v xf = v x 0
– Vertical motion under constant acceleration ( g )
Wednesday, Feb. 6, 2013
(
)
+ Spring
-9.8
t
= v1441-002,
v yf = v y0 + a y tPHYS
2013
y0
4
Projectile Motion
But the object is still moving!! Why?
Because it still has constant
Why?
velocity in horizontal direction!!
Maximum height
Wednesday, Feb. 6, 2013
Because there is no
acceleration in horizontal
direction!!
The
only acceleration in this
PHYS 1441-002, Spring 2013
Dr. Jaehoon
motion.
It isYua constant!!
5
Kinematic Equations for a projectile motion
x-component
ax = 0
vx = vxo
Dx = vxo t
y-component
a y = - g = -9.8 m s
v y = v yo - gt
Dy =
1
2
(v
)
+
v
t
yo
y
2
xo
v = v - 2gy
Dx = vxot
Dy = v yot - gt
v =v
2
x0
Wednesday, Feb. 6, 2013
2
y
2
yo
1
2
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
2
6
2
Show that a projectile motion is a parabola!!!
x-component
vxi = vi cos qi
v yi = vi sin q i
y-component
a =ax i + a y j = - gj
ax=0
x f = vxit = vi cosqi t
t=
xf
vi cos q 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 q it - gt
2
2
Plug t into
the above
ö
æ
ö 1 æ
xf
xf
y f = vi sin q i ç
÷ - 2 g ç v cos q ÷
è i
è vi cos q i ø
i ø
æ
ö 2
g
y f = x f tan q i - ç
xf
2
2
÷
2v
cos
q
è i
i ø
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
2
What kind of parabola is this?
7
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, Feb. 6, 2013
( )
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, Spring 2013
Dr. Jaehoon Yu
8
Ex.3.5 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, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
9
First, the initial velocity components
v0 = 22 m s
v0 y
q = 40
v0 x
(
)
sinq = ( 22m s ) sin 40
v0x = vo cosq = 22m s cos 40 = 17 m s
v0 y = vo
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
= 14 m s
10
Motion in y-direction is of the interest..
y
ay
vy
v0y
?
-9.8 m/s2
0 m/s
+14 m/s
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
t
11
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, Feb. 6, 2013
Solve for y
0 - 14 m s
)
2 -9.8m s
2
(
(
y=
2
)
v 2y - voy2
2a y
= +10 m
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
12
Ex.3.9 extended: The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
13
What is y when it reached the max range?
y
ay
0m
-9.80 m/s2
Wednesday, Feb. 6, 2013
vy
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
voy
t
14 m/s
?
14
Now solve the kinematic equations in y direction!!
y
ay
0
-9.80 m/s2
y = voy t + a yt
1
2
Two soultions
2
t 0
voy + a y t = 0
1
2
Wednesday, Feb. 6, 2013
vy
Since y=0
voy
t
14 m/s
?
(
0 = voy t + ayt = t voy + 12 a yt
2
1
2
)
or
Solve
for t
t
-voy
1
ay
2
=
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
-2voy
ay
=
-2 ×14
-9.8
= 2.9s
15
Ex.3.9 The Range of a Kickoff
Calculate the range R of the projectile.
(
)(
)
x  vox t + ax t = vox t = 17 m s 2.9 s = +49 m
1
2
Wednesday, Feb. 6, 2013
2
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
16
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 cosqi = 65.0 ´ cos37 = 51.9m / s
v yi = vi sinqi = 65.0 ´ sin37 = 39.1m / s
y f  -125.0 = v t - 1 gt 2 Becomes
yi
2
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
Since negative time does not exist.
t =Wednesday,
10.4s
Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
17
Example cont’d
• What is the speed of the stone just before it hits the ground?
vxf = vxi = vi cosqi = 65.0 ´ cos37 = 51.9m / s
v yf = v yi - gt = vi sinqi - gt =39.1- 9.80 ´10.4 = -62.8m / s
v =
v +v =
2
xf
2
yf
(
51.9 + -62.8
2
)
2
= 81.5m / s
• What are the maximum height and the maximum range of the stone?
Do these yourselves at home for fun!!!
Wednesday, Feb. 6, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
18