PHYS 1441 – Section 001
Lecture #6
Wednesday, June 17, 2015
Dr. Jaehoon Yu
•
•
•
•
Understanding the 2 Dimensional
Motion
2D Kinematic Equations of Motion
Projectile Motion
Maximum Range and Height
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
1
Announcements
• Quiz #2
–
–
–
–
Beginning of the class tomorrow, Thursday, June 18
Covers CH3.1 to what we finish today, Wednesday, June 17
Bring your calculator but DO NOT input formula into it!
You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam  no
solutions of any kind, derivations or word definitions!
• No additional formulae or values of constants will be provided!
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
2
Reminder: Special Project #2 for Extra Credit
• Show that the trajectory of a projectile motion
is a parabola and explain your answer.
– 20 points
– Due: Monday, June 22
– 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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
3
2D Displacement
Displacement:
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
4
2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
t - to
=
What is the direction of
the Average velocity?
Right! The same as the displacement!
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
5
2D Instantaneous Velocity
The instantaneous velocity indicates how fast the car
moves and the direction of motion at each instant of time.
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
6
2D Average Acceleration
t - to
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
=
7
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
How is each of
these quantities
defined in 1-D?
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
8
Kinematic Quantities in 1D and 2D
Quantities
1 Dimension
Displacement
Dx = x f - xi
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
Wednesday, June
17,
What
2015
2 Dimension
Dx x f - xi
vx =
=
Dt
t f - ti
Dx
vx º lim
Dt ®0 Dt
Dvx vxf - vxi
ax º
=
Dt
t f - ti
ax º lim
Dt®0
Dvx
Dt
PHYS 1441-001,
Summer
is the difference
between
1D 2015
and 2D quantities?
Dr. Jaehoon Yu
9
A Motion in 2 Dimension
This is a motion that could be viewed as two motions (x
and y directions) combined into one. (superposition…)
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
10
Motion in horizontal direction (x)
(v
+ vx ) t
vx = vxo + axt
Dx =
v = v + 2ax x
Dx = vxot + axt
2
x
2
xo
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
1
2
xo
1
2
2
11
Motion in vertical direction (y)
v y = v yo + a y t
Dy =
1
2
(v
)
+
v
t
yo
y
v = v + 2a y y
2
y
2
yo
Dy = v yot + a yt
1
2
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
12
2
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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
13
Kinematic Equations in 2-Dim
x-component
y-component
vx = vxo + axt
v y = v yo + a y t
Dx =
1
2
(v
+
v
t
)
xo
x
2
y
2
xo
Dx = vxot + ax t
1
2
Wednesday, June 17,
2015
1
2
(v
)
+
v
t
yo
y
v = v + 2a y y
v = v + 2ax x
2
x
Dy =
2
2
yo
Dy = v yot + ayt
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
1
2
14
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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
15
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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
16
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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
17
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 17,
2015
2
2
= +740 m
)(7.0 s) = +190m s
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
18
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 17,
2015
2
2
)(7.0 s)
2
= +390 m
)(7.0 s) = +98m s
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
19
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 17,
2015
2
m s
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
20
If we visualize the motion…
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
21
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 17, 2015
(
)
v y0 + Summer
-9.82015t
t = 1441-001,
v yf = v y0 + a yPHYS
22
Projectile Motion
But the object is still moving!! Why?
Because it still has constant
Why?
velocity in horizontal direction!!
Maximum height
Wednesday, June 17,
2015
Because there is no
acceleration in horizontal
direction!!
The
only acceleration in this
PHYS 1441-001, Summer 2015
Dr. Jaehoon
motion.
It isYua constant!!
23
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 17,
2015
2
y
2
yo
1
2
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
2
24
2
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 17,
2015
( )
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 2015
Dr. Jaehoon Yu
25
Ex.3.6 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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
26
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 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
= 14 m s
27
Motion in y-direction is of the interest..
y
ay
vy
v0y
?
-9.8 m/s2
0 m/s
+14 m/s
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
t
28
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 17,
2015
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 2015
Dr. Jaehoon Yu
29
Ex.3.6 extended: The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Wednesday, June 17,
2015
PHYS 1441-001, Summer 2015
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 17,
2015
vy
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
voy
t
14 m/s
?
31
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 17,
2015
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 2015
Dr. Jaehoon Yu
-2voy
ay
=
-2 ×14
-9.8
= 2.9s
32
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 17,
2015
2
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
33
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
June 17,
PHYS 1441-001, Summer 2015
2015
Dr. Jaehoon Yu
34
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, June 17,
2015
PHYS 1441-001, Summer 2015
Dr. Jaehoon Yu
35