PHYS 100: Lecture 4
PROJECTILE MOTION
v0
1.4
1.2
Velocity of Train
vT
y(m)
1
0.8
0.6
0.4
0.2
0
0
5
10
15
x(m)
y = (v0/vT) x – (g/2vT2)x2
Physics 100 Lecture 4, Slide 1
Music
Who is the Artist?
BB
A)
B)
C)
D)
E)
Miles Davis
Wynton Marsalis
Chris Botti
Nina Simone
Chet Baker
Why?
Valentine’s Day is coming !
“My Funny Valentine” by Chet is a classic !!!
DVD Must-See:
Twilight of his career/life
Van Morrison does “Send in the Clowns”
Physics 100 Lecture 4, Slide 2
WHAT DID YOU FIND DIFFICULT?
Im having a difficult time wrapping my head around the concept of relating the train
example to projectile motion. Everything is just a little confusing.
Relative Motion: Ball thrown straight up on train A. What is horizontal velocity of ball as
measured by a passenger on train B?
VAB = +30 m/s
e.g., VAG = +15
VBG = -15 VGB = +15
VAB = VAG + VGB = +30
(VballA)H = 0
(VballB)H = VAB
I did not completely understand what the formulas are identifying and how to use them to
get a specific variable.
substituting t =
xBG
vT
into yBG equation
yields an equation relating xBG and yBG
y BG =
v0
1 g 2
xBG − 2 xBG
vT
2 vT
“y vs x” and “y vs t” are parabolas
Question five has nothing to do with what we talked about in the prelecture.
THE BATTLESHIP QUESTION WAS THE HARDEST!!.. We will build towards it the
Physics 100 Lecture 4, Slide 3
entire lecture
THE BIG IDEAS
NOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE
1.
Motion described as 1-D freefall in one frame is described as
“projectile motion” in all other frames.
2. Description of projectile motion is superposition of x and y motions
a) Constant velocity in x (horizontal)
b) Constant acceleration in y (vertical)
Physics 100 Lecture 4, Slide 4
Ball on a Cart
DEMO: Ball is ejected straight up from a cup in a cart moving at constant velocity.
When it falls back , will it land (A) in back of the cup (B) in the cup (C) in front of the cup
A
v0
B
C
v
BB
This is THE POINT: Descriptions of same motion from two different reference frames
Cart frame: motion is 1-D free fall: ball goes up, stops, and comes straight back down
yC
v y = v0 − gt
y = v0t − 12 gt 2
vx = 0
x=0
xC
Ground frame: motion is 2-D projectile: ball follows parabolic trajectory
yG
v y = v0 − gt
y = v0t − 12 gt 2
vx = v
x = vt
xG
Physics 100 Lecture 4, Slide 5
Ball on a Cart II
DEMO: Ball is ejected straight up from a cup in a cart moving at constant acceleration.
When it falls back , will it land (A) in back of the cup (B) in the cup (C) in front of the cup
A
v0
B
C
a
BB
WHAT’S THE DIFFERENCE?
Ground frame: motion OF BALL is 2-D projectile: parabolic trajectory
motion OF BALL in x-direction is motion at constant velocity
motion OF CART in x-direction is motion with increasing velocity
xBALL = vt
xCART = vt + 12 at 2
Physics 100 Lecture 4, Slide 6
Preflight 1
Without air resistance, an object dropped from a plane flying at constant speed
in a straight line will:
BB
(A)
Quickly lag behind the plane
(C)
Move ahead of the plane
(B)
Remain vertically under the plane
(D)
It depends on the object
You said:
• The ball falls without any velocity in the x-direction,
so it will fall straight down as the plane continues to
move on at a constant velocity.
• Just as a ball thrown above a train. The flight of the
ball will maintain its horizontal x comp velocity unless
acted upon by another force.
Same as Ball on Cart !!
80
60
40
20
0
A
B
C
D
Physics 100 Lecture 4, Slide 7
Practice (with a Point)
Two objects are thrown straight up from the same height. Object 1 is given an
initial velocity v0 while Object 2 is given an initial velocity 2v0.
BB
Which object is in the air the longest?
(A) Object 1
ttop
(B) Object 2
(C) Same amount of time
2v0
v
= 0
g
ttop =
v0
2v0
g
2
1
v1 = v0 − gt
v2 = 2v0 − gt
• How do we calculate how long the ball stays in the air?
• At the top (1/2 of the time in the air) the velocity of the ball is zero
• The velocity is being reduced by g m/sec every second
2v0
v
v0
1
2
t
Physics 100 Lecture 4, Slide 8
More Practice
Two objects are thrown with the same initial velocity from the same height.
Object 1 is thrown at angle θ = 45o while Object 2 is thrown at angle θ = 60o
BB
Which object is in the air the longest?
(A) Object 1
ttop =
(B) Object 2
v0 sin 45°
g
(C) Same amount of time
v0
v0
45o
ttop =
60o
1
v0 sin 60°
g
2
(v2 ) y = v0 sin 60° − gt
(v1 ) y = v0 sin 45° − gt
• How do we calculate how long the ball stays in the air?
• At the top (1/2 of the time in the air) the y-component of the velocity of
the ball is zero
• The y-component of the velocity is being reduced by g m/sec every second
vy
v0sin60o
v0sin45o
1 2
t
Physics 100 Lecture 4, Slide 9
More Practice
Two objects are thrown with different initial velocities from the same height. Object 1
is thrown with v0 at angle θ = 60o while Object 2 is thrown with 1.5v0 at angle θ = 30o
Which object is in the air the longest?
(A) Object 1
ttop
(B) Object 2
v sin 60°
= 0
g
(C) Same amount of time
v0
1.5v0
60o
ttop =
30o
1
1.5v0 sin 30°
g
BB
2
(v2 ) y = 1.5v0 sin 30° − gt
(v1 ) y = v0 sin 60° − gt
• How do we calculate how long the ball stays in the air?
• At the top (1/2 of the time in the air) the y-component of the velocity of
the ball is zero
• The y-component of the velocity is being reduced by g m/sec every second
vy
Sin60o = √3/2 = .87
Sin30o = 1/2 = .50
v0sin60o
1.5v0sin30o
2
√3
60o
1
1.5 Sin30o = 3/4 = .75
2
1
t
Physics 100 Lecture 4, Slide 10
The Point
What determines how long an object will stay in the air when thrown?
(A)
(B)
(C)
(D)
Only the magnitude of the initial velocity
Only the horizontal component of the initial velocity
Only the vertical component of the initial velocity
Both the horizontal & vertical component of the initial velocity
BB
• How do we calculate how long the ball stays in the air?
• At the top (1/2 of the time in the air) the y-component of the velocity of
the ball is zero
• The y-component of the velocity is being reduced by g m/sec every second
v y = v0 sin θ − gt
ttop =
v0 sin θ
g
Physics 100 Lecture 4, Slide 11
Preflight 3
A man and his son are standing at the edge of the cliff. They both pick up very similar
rocks and throw them horizontally off of the cliff. The man’s rock travels twice as
far from the base of the cliff as the rock the boy threw. If they both threw their
rocks at the same time and from the same height, which one hit the ground first?
(A)
The boy’s rock
(C)
They land at the same time
(B)
The man’s rock
(D)
Need more information
You said:
• For the boy's rock to go a shorter distance, it had to
have a slower velocity. Distance is the velocity
multiplied by time. Therefore, since the man's rock
traveled further, it had to have a greater velocity and
time.
• Since they both throw the rocks horizontally off of
the cliff, we know that the velocities of the rocks have
no y-component. Thus, the only thing causing the rocks
to have motion in the y-direction is the acceleration due
to the force of gravity. Since this acceleration is the
same for both rocks, both will reach the ground at the
same time.
DEMO
BB
60
50
40
30
20
10
0
A
B
C
D
Physics 100 Lecture 4, Slide 12
How High?
Two objects are thrown with different initial velocities from the same height. Object 1
is thrown with v0 at angle θ = 60o while Object 2 is thrown with 1.5v0 at angle θ = 30o
Which object goes the highest?
(A) Object 1
ttop
(B) Object 2
v sin 60°
= 0
g
(C) Same amount of time
v0
1.5v0
60o
30o
1
ttop =
1.5v0 sin 30°
g
BB
2
(v2 ) y = 1.5v0 sin 30° − gt
(v1 ) y = v0 sin 60° − gt
• How do we calculate how high the ball goes?
• At the top (1/2 of the time in the air) the y-component of the velocity of
the ball is zero
• Find the time when this occurs (we’ve done that a lot today)
• Determine the height at this time.
THE MATH:
y = v0 sin θt − 12 gt 2
v sin θ
t= 0
g
ymax
v sin θ 1  v0 sin θ 

= v0 sin θ 0
− 2 g 
g
 g 
2
v02 sin 2 θ
1
ymax =
2
g
Physics 100 Lecture 4, Slide 13
Preflight 5
A battleship simultaneously fires two
shells at enemy ships. If the shells
follow the parabolic trajectories
shown, which ship gets hit first?
(A) A
(B) At the same time
(C) B
(D) more info
You said:
• Because boat A is closer, but has a taller trajectory, it
will hit at the same time as boat B, which has a flatter
trajectory but is further away.
BB
• A goes at a higher height and as a result it will stay in
the air longer than B.
• We do not have much information for this question
because we do not know which angle the shells were
thrown and how much distance the shells went for A
and B.
50
40
30
20
10
0
A
B
C
D
Physics 100 Lecture 4, Slide 14
Preflight 5
A battleship simultaneously fires two
shells at enemy ships. If the shells
follow the parabolic trajectories
shown, which ship gets hit first?
(A) A
(B) At the same time
(C) B
(D) more info
We certainly have not been given the velocities and the distances
However, is there something about this picture that tells us enough
to be able to answer the question ????
YOU BETCHA !!!
The shell to A goes higher which means its initial vertical component of
the velocity is greater than that of the shell that went to B.
We can use this fact to compare the time in the air of the shells.
Physics 100 Lecture 4, Slide 15
Preflight 5
A battleship simultaneously fires two
shells at enemy ships. If the shells
follow the parabolic trajectories
shown, which ship gets hit first?
(A) A
(B) At the same time
(C) B
(D) more info
WHAT HAVE WE LEARNED ABOUT PROJECTILE MOTION TODAY ??
ttop
ymax
v sin θ
= 0
g
The time in the air is determined by the
vertical component of the initial velocity
v02 sin 2 θ
=−
g
The maximum height is determined by the
vertical component of the initial velocity
1
2
A goes HIGHER so A takes longer !!
Physics 100 Lecture 4, Slide 16
Physics 100 Lecture 4, Slide 17
A Comment
not seen in Fall 10 !!
I think we should address concerns expressed in comments like these:
“I found the many equations shown on the pre-lecture to be confusing. Is there
an simpler way to address situations without relying on equations? “
“all of the equations with nothing but variables, i need some examples with
numbers to understand better what is being done it is hard to follow what is
happening to letters. and all of the subscripts get confusing”
“Symbolic relationships with projectile motion--that is, "this increases with the
square of that" without actually having numbers to look at.”
My take on this:
Using symbols is ESSENTIAL for success in PHYS 211
All equations are not created equal; there are only a few that are always true
Math is used as a thinking tool in physics
Arguments are made with symbols, not numbers
Subscripts help clarify meaning of variables
All of this takes practice: We will be focusing on these issues when we address
problem solving in the second half of the semester
Physics 100 Lecture 4, Slide 18