Motion in Two Dimensions

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Motionin Two Dimensions

• Chapter 7.2

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Projectile Motion

• What is the path of a projectile as it moves
  through the air?
• Parabolic?
• Straight up and down?
• Yes, both are possible.
• What forces act on projectiles?
• Only gravity, which acts only in the negative
  y-direction.
• Air resistance is ignored in projectile motion.

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Choosing Coordinates Strategy

• For projectile motion
• Choose the y-axis for vertical motion where
  gravity is a factor.
• Choose the x-axis for horizontal motion. Since
  there are no forces acting in this direction (of
  course we will neglect friction due to air
  resistance), the speed will be constant (a 0).
• Analyze motion along the y-axis separate from the
  x-axis.
• If you solve for time in one direction, you
  automatically solve for time in the other
  direction.

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The Trajectory of a Projectile

• What does the free-body diagram look like for
  force?

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The Vectors of Projectile Motion

• What vectors exist in projectile motion?
• Velocity in both the x and y directions.
• Acceleration in the y direction only.

vx (constant)
ax 0
vy (Increasing)
Trajectory or Path

• Why is the velocity constant in the x-direction?
• No force acting on it.

• Why does the velocity increase in the
  y-direction?
• Gravity.

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Formulas for Motion of Objectsassuming d is
displacement from origin and time starts at 0.
Equations to use when an accelerating object has an initial velocity. Form to use when accelerating object starts from rest (vi 0).
d ½ (vi vf) t vavet d ½ vf t
vf vi at vf at
d vi t ½ a(t)2 d ½ a(t)2
vf2 vi2 2ad vf2 2ad
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There are only so many things a projectile can do
Go up and down
Be shot at an angle
Fall
d vi t ½ at2 vf vi at vf 2vi
22ad Beware cases where t ltgt hang time
Solve x and y separately. t is when the object
hits the ground.
d ½ at2 vf vi at vf 2vi 22ad Note viltgt0
is possible
Move horizontally
Be shot horizontally
Be shot at an angle
d vave t vf vi at vave ½(vfvi) vf 2vi
22ad
Solve x and y separately. t is when the object
hits the ground.
Solve x and y separately. t is when the object
stops.
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Ex. 1 Launching a Projectile Horizontally

• A cannonball is shot horizontally off a cliff
  with an initial velocity of 30 m/s. If the
  height of the cliff is 50 m
• How far from the base of the cliff does the
  cannonball hit the ground?
• With what speed does the cannonball hit the
  ground?
• Note This is the same problem as a life raft
  is dropped from an airplane

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Diagram the problem
vi
Fg Fnet
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State the Known Unknown

• Known
• xi 0
• vix 30 m/s
• yi 0
• viy 0 m/s
• a -g
• y -50 m
• Unknown
• x at y -50 m
• vf ?

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Perform Calculations (y)

• y-direction
• vy -gt
• y viyt ½ gt2
• Using the first formula above
• vy (-9.8 m/s2)(3.2 s) 31 m/s

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Perform Calculations (x)

• x-Direction
• x vixt
• x (30 m/s)(3.2 s) 96 m from the base.
• Using the Pythagorean Theorem
• v vx2 vy2
• v (30 m/s)2 (31 m/s)2 43 m/s

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Ex. 2 Projectile Motion above the Horizontal

• A ball is thrown from the top of the Science Wing
  with a velocity of 15 m/s at an angle of 50
  degrees above the horizontal.
• What are the x and y components of the initial
  velocity?
• What is the balls maximum height?
• If the height of the Science wing is 12 m, where
  will the ball land?

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Diagram the problem
Fg Fnet
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State the Known Unknown
Hor Vert
Vix Viy
Vfx Vfy
dx dy
ax ay
t t
?

• Known
• xi
• yi
• vi
• ?
• a
• Unknown
• ymax ?
• t ?
• x ?
• viy ?
• vix ?

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State the Known Unknown

• Known
• xi 0
• yi 12 m
• vi 15 m/s
• ? 50
• a -g
• Unknown
• ymax ?
• t ?
• x ?
• viy ?
• vix ?

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Perform the Calculations (ymax)

• y-direction
• Initial velocity viy visin?
• viy (15 m/s)(sin 50)
• viy 11.5 m/s
• Time when vfy 0 m/s vfy viy at viy gt
• 0 viy gt ? viy gt ? t viy / g
• t (11.5 m/s)/(9.81 m/s2)
• t 1.17 s
• Determine the maximum height d vit ½ at2
• dmax viyt ½ gt2, plus 12 m
• ymax 12 m (11.5 m/s)(1.17 s) ½ (9.81
  m/s2)(1.17 s)2
• ymax 18.7 m

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Perform the Calculations (t)

• Since the ball will accelerate due to gravity
  over the distance it is falling back to the
  ground, the time for this segment can be
  determined as follows
• Time when ball hits the ground ymax viyt ½
  gt2
• Since yi can be set to zero as can viy,
• t 2ymax/g
• t 2(18.7 m)/ (9.81 m/s2)
• t 1.95 s
• By adding the time it takes the ball to reach its
  maximum height to the time it it takes to reach
  the ground will give you the total time.
• ttotal 1.17 s 1.95 s 3.12 s

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Perform the Calculations (x)

• x-direction
• Initial velocity vix vicos?
• vix (15 m/s)(cos 50)
• vix 9.64 m/s
• Determine the total distance x vixt
• x (9.64 m/s)(3.12 s)
• x 30.1 m

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Analyzing Motion in the x and y directions
independently.

• x-direction
• dx vix t vfxt
• vix vi?cos?
• y-direction
• dy ½ (vi vf) t vavg t
• vf viy gt
• dy viy t ½ g(t)2
• vfy2 viy2 2gd
• viy vi?sin?

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Key Ideas

• Projectile Motion
• Gravity is the only force acting on a projectile.
• Choose a coordinate axis that where the
  x-direction is along the horizontal and the
  y-direction is vertical.
• Solve the x and y components separately.
• If time is found for one dimension, it is also
  known for the other dimension.