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Linear Motion

v ≡

d�t

a ≡

Δv�t

d = v·t

Definition of Average Velocity�(The rate POSITION CHANGES)

Definition of Acceleration�(The rate VELOCITY CHANGES)

Instantaneous Final Velocity�when Accelerating�(How FAST?)

Distance Traveled�when Accelerating�(How FAR?)

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Projectile�Motion

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Clifton Bluhm

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Know the key concept behind Two Dimensional Motion .

Be prepared to do theBull’s Eye LABNext Block Day!!!

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I can . . .

  • explain the behavior of the vertical and horizontal components of projectile motion.
  • accurately predict the position of a projectile at any point in time, given initial conditions.

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Which Bullet Lands First?

Dropped or Shot

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Which Bullet Lands First?

Dropped or Shot

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Which Bullet Lands First?

Dropped or Shot

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Ball Dropped Versus Ball Shot

Demonstration

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Horizontal motion is CONSTANT!

Vertical motion is ACCELERATING

y

y

y

y

y

y

y

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No Gravity, But Does Have Horizontal Velocity

velocity is same everywhere!�d = v · t

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y

y

y

y

y

y

y

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Gravity, AND Horizontal Velocity

.

Projectile Motion

Throw a ball into the air!

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Gravity, AND Horizontal Velocity

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Gravity, AND Horizontal Velocity

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POSITION

y

y

y

y

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VELOCITY

y

y

y

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VELOCITY

V

Vy0

Vx0

AP Physics 1

y

y

y

y

y

y

y

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y

x

y0

x0

vy0

vx0

v

θ

vx = constant

x = x0 + vx0 ∙t

Horizontal

vy = vy0 + g∙t

y = y0 + vy0 ∙t + ½g∙t2

Vertical

(x,y)

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y

x

y0

x0

vy0

vxi

v

θ

vyo

vxo

vo

Honors Physics

vx = constant

x = xo + vxo∙t

Horizontal

vy = vyo + g∙t

y = yo + vyo∙t + ½g∙t2

Vertical

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?

10.0 m

2.5 m

vx = 1.8 m/s

dy = ½ a · t2

10.0 = ½ 9.80 · t2

t = 1.41 seconds

dx = vx · t

2.5 = vx · 1.41

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If the bridge is 18.0 meters off of the water, and the car is traveling at 32.0 m/s, how far horizontally will the car travel before it hits the water?

18.0 m

?

0 = 18.0 + (0)t + ½(-9.80)t2

t = 1.92 seconds

x = 0 + 32.0·1.92

x = 61.3 meters

y

y

y

y

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0 = 14.0 + (0)t + ½(-9.80)t2

t = 1.69 seconds

x = 0 + 27.0·1.69

x = 45.6 meters

y

y

y

y

?

14.0 m

27.0 m/s

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http://www.siue.edu/ALUMNI/alumni_target.html

58.0 m

Vx = 175 m/s

dy = ½ a · t2

dy = ½(9.80)(.331)2

t = 58.0 / 175

t = .331 s

dx = vx · t

t = dx / vx

dy = .538 m

dy = .538 m

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37°

50 m/s

d = ½·a·t2

d = ½·10·t2

t d

1 s 5 m

2 s 20 m

3 s 45 m

4 s 80 m

5 s 125 m

40 m/s

30 m/s

How far will the object land downrange?

d = 40 m/s · 6 s

d = 240 m

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http://petsports.vstore.ca/product_info.php/products_id/164

http://www.siue.edu/ALUMNI/alumni_target.html

(58.0, y)

175 m/s

82.158

154.52

y

y

y

y

58.0 = 0 + 154.52·t

y = 0+ 82.158(.375) + ½(-9.80)(.375)2

t = .375 seconds

y = 30.14 meters

28°

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10 ft

40 ft

A kangaroo can jump 40 ft horizontally and 10 ft vertically. How fast is it running?

dy = ½ a · t2

10 = ½(32.2)(t)2

t = .79 s

tair= 2(.79) = 1.6 s

dx = vx · t

40 = vx · 1.6

vx = 25 ft/s

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Robbie Maddison – 120 ft Arc de Triomphe, Las Vegas 2008

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Robbie Maddison – 120 ft Arc de Triomphe 2008

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3.27 seconds of AIR!

322 ft (98.1 meters)

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28.1°

34.0 m/s

30.0 m/s

16.0 m/s

Vy = a·t

Vy = 9.80·1.64

tup = 1.64 s

ttotal = 3.27 s

dx = Vx· t

Vx = 30.0 m/s

98.1 = Vx · 3.27

98.1 meters

76.1 mi/hr

Vy = 16.0 m/s

V = 16.02 + 30.02

V = 34.0 m/s

?

?

tan(?) = 16.0/30.0

? = 28.1

dy= ½ a·t2

dy= ½ 9.80·1.642

dy= 13.1 m

dy= 43.1 ft

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d = ½ a·t2

= ½ 9.80·(1 sec)2

LASER

4.90 m

8 km

= 4.90 m

Light

Amplification by the

Stimulated

Emission of

Radiation

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4.90 m

?

? = 7,901 m

6,371,020 m

6,371,020 m

? = √(6,371,020 + 4.90)2 – 6,371,0202

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30 m/s

30 m/s

28.2 m/s

28.2 m/s

10.3 m/s

10.3 m/s

20°

70°

Hang Time = 2.09 seconds

Hang Time = 5.64 seconds

dx = Vx · t

dx = 10.3 · 5.64

dx = 58.1 m

dx = Vx · t

dx = 58.1 m

dx = 28.2 · 2.06

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V0 m/s

V0cos(θ)

V0sin(θ)

θ

V0 m/s

V0sin(90-θ)

V0cos(90- θ)

(90- θ)

-V0sin(θ) = V0sin(θ) - 9.8∙t

-2V0sin(θ) = -9.8∙t

2∙V0sin(θ) / 9.8 = t

x = 0 + V0cos(θ)∙t

x = V0cos(θ)∙ 2V0sin(θ) / 9.8

x = 2∙V02cos(θ)sin(θ) / 9.8

x = 2∙V02cos(90-θ)sin(90-θ) / 9.8

x = 2∙V02sin(θ)cos(θ) / 9.8

cos(90-θ) = sin(θ)

sin(90-θ) = cos(θ)

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http://www.novanewsnow.com/imgs/dynamique/articles/gros/track.jpg

45°

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http://4.bp.blogspot.com/_9T2cRtlgHV0/SKrMTJYz1UI/AAAAAAAAAls/Q0LQAdwYzkk/s400/Women%27s-Javelin-Throw-Mikaela-Ingberg.jpg

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http://www.iaaf.net/mm/photo/competitions/competition/04/77/03/47703_full-lnd.jpg

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I can . . .

  • explain the relationship between the vertical and horizontal components of projectile motion.
  • accurately predict the position of a projectile at any point in time, given initial conditions.