Chapter 3 – 2D MotionChapter 3 – 2D Motion

I.I. DefinitionsDefinitions

II.II. Projectile motionProjectile motion

Find the horizontal and vertical components of the d = 140 m displacement of a superhero who flies from the top of a tall building following the path shown in the Figure where = 30.0°.

Superperson

2 Dimensional Motion

• We will consider motion the the x-y plane.

• Positions now have (x,y) coordinates so we need to use vectors.

• There are two types of problems we need to consider– Throw or drop an object at an angle to the

horizontal– Make something go around in a circle

Position vector:Position vector: extends from the origin of a coordinate system to the particle. extends from the origin of a coordinate system to the particle.

jyixr ˆˆ

jyyixxrrr ˆ)(ˆ)( 121212

I.I. DefinitionsDefinitions

Average velocity:

jt

yit

x

t

rvavg ˆˆ

Displacement vector: Displacement vector: represents a particle’s position change during a certainrepresents a particle’s position change during a certain time interval.time interval.

Instantaneous velocity:Instantaneous velocity:

jdt

dyi

dt

dx

dt

rdkvjvivv zyx

ˆˆˆˆˆ

-The direction of the instantaneous velocity of a The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at particle is always tangent to the particle’s path at the particle’s positionthe particle’s position

Instantaneous accelerationInstantaneous acceleration:

Average acceleration:Average acceleration:t

v

t

vvaavg

12

jdt

dvi

dt

dv

dt

vdjaiaa yx

yxˆˆˆˆ

Instantaneous velocity:Instantaneous velocity:

jdt

dyi

dt

dx

dt

rdjvivv yx

ˆˆˆˆ

-The direction of the instantaneous velocity of a The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at particle is always tangent to the particle’s path at the particle’s positionthe particle’s position

Instantaneous accelerationInstantaneous acceleration:

Average acceleration:Average acceleration:t

v

t

vvaavg

12

jdt

dvi

dt

dv

dt

vdjaiaa yx

yxˆˆˆˆ

Projectile Motion

RANGE (R)

height (h)

What is its velocity here?It’s acceleration?How long did it take to get here?

here?

II. Projectile motionII. Projectile motion

Motion of a particle launched with initial velocity, vMotion of a particle launched with initial velocity, v00 and free fall acceleration and free fall acceleration

g.g.

- - Horizontal motion:Horizontal motion: aaxx=0 =0 v vxx=v=v0x0x= cte= cte

- - Vertical motion:Vertical motion: aayy= -g = -g

Range (R):Range (R): horizontal distance traveled by a horizontal distance traveled by a projectile before returning to launch height.projectile before returning to launch height.

tvtvxx x )cos( 0000

200

200 2

1)sin(

2

1gttvgttvyy y

gtvvy 00 sin

The horizontal and vertical motions are independent from each other.The horizontal and vertical motions are independent from each other.

- Trajectory: - Trajectory: projectile’s path.projectile’s path.

200

2

0

2

000000

00

)cos(2)(tan

cos2

1

cossin

cos

v

gxxy

v

xg

v

xvy

v

xt

000 yx

- Horizontal range: - Horizontal range: R = x-x R = x-x00; y-y; y-y00=0.=0.

0

202

000

022

0

2

0

2

000000

200

0000

2sincossin2

cos2

1tan

cos2

1

cos)sin(

2

1)sin(0

cos)cos(

g

vv

gR

v

RgR

v

Rg

v

Rvgttv

v

RttvR

(Maximum for a launch angle of 45º )

Overall assumption:Overall assumption: the air through which the projectile moves has no effect the air through which the projectile moves has no effect on its motion on its motion friction neglected. friction neglected.

In Galileo’s Two New Sciences, the author states that “for elevations (angles of In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” º by equal amounts, the ranges are equal…” Prove this statement.Prove this statement.

45

45

45

2

1

290sin452sin'

290sin452sin

20

20

20

20

g

v

g

vR

g

v

g

vR

x

v0

x=R=R’?

y

θ=45º

02sin: max0

20 hatdg

vRRange

bababa

bababa

sincoscossin)sin(

sincoscossin)sin(

)2cos()2sin(90cos)2cos(90sin'

)2cos()2sin(90cos)2cos(90sin

20

20

20

20

g

v

g

vR

g

v

g

vR