PHYS 1441 – Section 001
Lecture #6
Wednesday, June 11, 2014
Dr. Jaehoon Yu
•
•
•
Wednesday, June 11,
2014
Understanding the 2 Dimensional
Motion
2D Kinematic Equations of Motion
Projectile Motion
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
Announcements
• Reading Assignment
– CH3.7
• Quiz #2
– Beginning of the class tomorrow, Thursday, June, 12
– Covers CH3.1 to what we finish today, Wednesday, June 11
– Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!
– You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam  no
solutions, derivations or definitions!
• No additional formulae or values of constants will be provided!
• 1st Term exam results
– Class average: 61.2/99
• Equivalent to 61.8/100
– Top score: 99/99
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
Reminder: Special Project #2
• Show that the trajectory of a projectile motion
is a parabola!!
– 20 points
– Due: Monday, June 16
– You MUST show full details of your OWN
computations to obtain any credit
• Beyond what was covered in this lecture note
and in the book!
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
3
A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one. (superposition…)
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
4
Motion in horizontal direction (x)
(v
)
vx = vxo + axt
x=
v = v + 2ax x
x = vxot + ax t
2
x
2
xo
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
2
+
v
t
xo
x
1
2
2
5
Motion in vertical direction (y)
v y = v yo + a y t
y=
1
2
(v
)
+
v
t
yo
y
v = v + 2a y y
2
y
2
yo
y = v yot + a y t
1
2
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
6
A Motion in 2 Dimension
Imagine you add the two 1 dimensional motions on the left.
It would make up a one 2 dimensional motion on the right.
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
7
Kinematic Equations in 2-Dim
x-component
y-component
vx = vxo + axt
v y = v yo + a y t
x=
1
2
(v
)
+
v
t
xo
x
2
y
2
xo
Dx = vxot + ax t
1
2
Wednesday, June 11,
2014
(v
)
+
v
t
yo
y
v = v + 2a y y
v = v + 2ax x
2
x
y=
1
2
2
2
yo
Dy = v yot + ayt
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
2
8
2
Ex. A Moving Spacecraft
In the x direction, the spacecraft in zero-gravity zone has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x
and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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How do we solve this problem?
1. Visualize the problem  Draw a picture!
2. Decide which directions are to be called positive (+) and
negative (-). Normal convention is the right-hand rule.
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least
three of the kinematic variables. Do this for x and y
separately. Select the appropriate equation.
5. When the motion is divided into segments, remember
that the final velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
10
Ex. continued
In the x direction, the spacecraft in a zero gravity zone has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x
and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22.0 m/s
7.0 s
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14.0 m/s
7.0 s
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
11
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
Dx = voxt + axt
(
1
2
)(
2
) ( 24m s )(7.0 s)
= 22m s 7.0 s +
2
1
2
vx = vox + axt
(
) (
= 22m s + 24m s
Wednesday, June 11,
2014
2
2
= +740 m
)(7.0 s) = +190m s
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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Now, the motion in y-direction…
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14 m/s
7.0 s
Dy = voy t + a y t
(
)(
1
2
2
) (
= 14m s 7.0 s + 12m s
1
2
v y  voy + a y t
(
) (
= 14m s + 12m s
Wednesday, June 11,
2014
2
2
)(7.0 s)
2
= +390 m
)(7.0 s) = +98m s
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
13
The final velocity…
v
v y = 98m s
q
vx = 190 m s
(
) + ( 98m s)
(98 190) = 27
v  190 m s
q = tan
-1
2
= 210 m s
A vector can be fully described when
the magnitude and the direction are
given. Any other way to describe it?
Yes, you are right! Using
components and unit vectors!!
Wednesday, June 11,
2014
2
m s
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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If we visualize the motion…
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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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, June 11, 2014
(
)
v y0 + Summer
-9.82014t
t = 1441-001,
v yf = v y0 + a yPHYS
16
Projectile Motion
But the object is still moving!! Why?
Because it still has constant
Why?
velocity in horizontal direction!!
Maximum height
Wednesday, June 11,
2014
Because there is no
acceleration in horizontal
direction!!
The
only acceleration in this
PHYS 1441-001, Summer 2014
Dr. Jaehoon
motion.
It isYua constant!!
17
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, June 11,
2014
2
y
2
yo
1
2
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
18
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, June 11, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
What kind of parabola is this?
19
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, June 11,
2014
( )
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 1441-001, Summer 2014
Dr. Jaehoon Yu
20
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, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
21
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, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
= 14 m s
22
Motion in y-direction is of the interest..
y
ay
vy
v0y
?
-9.8 m/s2
0 m/s
+14 m/s
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
t
23
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, June 11,
2014
y=
Solve for y
0 - 14 m s
)
2 -9.8m s
2
(
(
2
)
v 2y - voy2
2a y
= +10 m
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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Ex.3.9 extended: The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Wednesday, June 11,
2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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What is y when it reached the max range?
y
ay
0m
-9.80 m/s2
Wednesday, June 11,
2014
vy
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
voy
t
14 m/s
?
26
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, June 11,
2014
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 1441-001, Summer 2014
Dr. Jaehoon Yu
-2voy
ay
=
-2 ×14
-9.8
= 2.9s
27
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, June 11,
2014
2
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
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