phys 100: lecture 4 - course websites · phys 100: lecture 4 ... – these problems help you...

Physics 100 Lecture 4, Slide 1

PHYS 100: Lecture 4

PROJECTILE MOTION

0

0.2

0.4

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0.8

1

1.2

1.4

0 5 10 15

x(m)y(m

)

y = (v0/vT) x – (g/2vT2)x2

Velocity of TrainvT

v0


Music

Who is the Artist?

A) Miles DavisB) Wynton MarsalisC) Chris BottiD) Nina SimoneE) Chet Baker

West Coast “cool” jazz from the 50’sIn honor of Valentine’s Day next week:

His definitive version of My Funny ValentineHypnotizing voice….

DVD Must-See:Twilight of his career/lifeVan Morrison does “Send in the Clowns”

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POINTS POINTS POINTSUNDERSTANDING is the GOAL; POINTS are what we give to get you to do the work

What are the takeaway messages from this graph?• A and B students are indistinguishable on EFFORT (>95% avg)• C and D students do not put in the same effort (70-80%)• A and B students are distinguished by QUIZ and EXAM grades• C and D students do poorly on exams (esp quizzes)

Phys 100 Fall 2010 Data

EFFORT

DISCHWLECT (PL,CP,LECT)

PERFORMANCE

QUIZEXAM (MIDTERM)FINAL

NOTE: Performance on Quizzes is your best indicator of your CURRENT understanding


How To Do Homeworks• Interactive Examples

– These problems help you develop a “concept-based” problem solving approach

REPEAT OF SLIDE FROM LEC 1

– You get another question to help you develop your Plan !– Eventually, we “give away the store”: This is an “Example” !!!

• If you can’t calculate the answer to the problem, select the Help Button


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 motionsa) Constant velocity in x (horizontal)b) Constant acceleration in y (vertical)

THE BIG IDEASNOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE


1. The Plan is to build to an understanding of the Battleship CheckPoint2. How will we do this?

a) We will do clicker questions whose explanations will feature equations

b) The purpose is not to memorize these equations but to see how you can REASON with equations !!!

THE PLAN FOR TODAY


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

vB

A C

This is THE POINT: Descriptions of same motion from two different reference frames

gtvvy −=0

0=xv

v0

2

2

10

gttvy −=

0=x

Cart frame: motion is 1-D free fall: ball goes up, stops, and comes straight back down

xC

yC

gtvvy −=0

vvx =

2

2

10

gttvy −=

vtx =

Ground frame: motion is 2-D projectile: ball follows parabolic trajectory

xG

yG

BB


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

aB

A C

WHAT’S THE DIFFERENCE?

v0

Ground frame: motion OF BALL is 2-D projectile: parabolic trajectorymotion OF BALL in x-direction is motion at constant velocitymotion OF CART in x-direction is motion with increasing velocity

2

2

1 atvtxCART +=vtxBALL =

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CheckPoint 1Without air resistance, an object dropped from a plane flying at constant speed

in a straight line will:

You said:

• quickly lag behind the plane because after it has been dropped it no longer has the same velocity as the plane and it is going in a different direction also

• This is just like the boat moving at a constant speed and someone from the boat throwing it straight up and the ball would fall straight down. So the object dropped from a plane flying will fall straight down, which is vertically under the plane.

(A) Quickly lag behind the plane

(B) Remain vertically under the plane

(C) Move ahead of the plane

(D) It depends on the object

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40

60

80

A B C D

Same as Ball on Cart !!

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Practice (with a Point)

• 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

gtvv −= 01

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.

Which object is in the air the longest?

(A) Object 1 (B) Object 2 (C) Same amount of time

v0

2v0

1 2

gtvv −= 02 2

g

vttop

0=

g

vttop

02=

v

t

v0

2v0

1 2

BB


More Practice

• 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

gtvv y −°= 45sin)(01

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



v0 v0

1 2

g

vttop

°=

45sin0

vy

t

v0sin60o

1 2

60o45o

v0sin45o


g

vttop

°=

60sin0

BB


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



v0 1.5v0

1 230o60o


g

vttop

°=

60sin0

gtvv y −°= 30sin5.1)(02

g

vttop

°=

30sin5.10

60o

1

2√3

Sin60o = √3/2 = .87Sin30o = 1/2 = .50

1.5 Sin30o = 3/4 = .75

1.5v0sin30o

vy

t

v0sin60o

12

BB


The PointWhat determines how long an object will stay in the air when thrown?

(A) Only the magnitude of the initial velocity(B)(C)

Both the horizontal & vertical component of the initial velocity (D)

Only the horizontal component of the initial velocityOnly the vertical component of the initial velocity


gtvvy −= θsin0

g

vttop

θsin0

=

BB


CheckPoint 2

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20

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80

A B C D

You said:

• The boys rock would hit first because it did not go as far. The mans rock went farther and so it will take longer to hit the ground.

• The y component (which is gravity) is the same, therefore, since their initial position on the y-axis is identical, their fall will happen at the same place and ultimately result in the rocks landing simultaneously.

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

(B) The man’s rock

(C) They land at the same time

(D) Need more information

DEMO

BB


How High?

• 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.

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?


v0 1.5v0

1 230o60o


g

vttop

°=

60sin0

gtvv y −°= 30sin5.1)(02

g

vttop

°=

30sin5.10

THE MATH:2

2

10sin gttvy −= θ

g

vt

θsin0=

2

0

2

10

0max

sinsinsin

−=

g

vg

g

vvy

θθθ

g

vy

θ22

0

21

maxsin

=

BB


CheckPoint 3

A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first?

You said:

• The trajectory of the missile is different for each shot but the actual distance that each missile travels is the same. This means that the missile will be traveling the same distance with the same velocity making the time of each missiles flight the same.

• It hits B first because no matter how fast the shells are traveling gravity pulls the sames for both and so the one that does not have to travel as high will hit its target first.

• In order to answer this, I need to know the angle of the trajectories and the distance each ship is away from the battleship.

(A) A

(D) more info

(C) B

(B) At the same time

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A B C D

BB


CheckPoint 3


We certainly have not been given the velocities and the distances

(A) A

(D) more info

(C) B


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.


CheckPoint 3


WHAT HAVE WE LEARNED ABOUT PROJECTILE MOTION TODAY ??

(A) A

(D) more info

(C) B


g

vttop

θsin0=

The time in the air is determined by the vertical component of the initial velocity

g

vy

θ22

0

21

max

sin−=

The maximum height is determined by the vertical component of the initial velocity

A goes HIGHER so A takes longer !!

phys 100: lecture 4 - course websites · phys 100: lecture 4 ... – these problems help you...

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