CHAPTER 6 MOTION IN TWO DIMENSIONS

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CHAPTER 6MOTION IN TWO DIMENSIONS

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• In this chapter you will
•  
•  Use Newtons laws and your knowledge of vectors
  to analyze motion in two dimensions.
• Solve problems dealing with projectile and
  circular motion.
• Solve relative velocity.

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CHAPTER 6 SECTIONS

• Section 6.1 Projectile Motion
• Section 6.2 Circular Motion
• Section 6.3 Relative Velocity

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SECTION 6.1 PROJECTILE MOTION

• Objectives
• Recognize that the vertical and horizontal
  motions of a projectile are independent.
• Relate the height, time in the air, and initial
  vertical velocity of a projectile using its
  vertical motion, and then determine the range
  using the horizontal motion.
• Explain how the trajectory of a projectile
  depends upon the frame of reference from which it
  is observed.

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INTRO

• Parabola set of all points equidistant from a
  fixed line called the directix, and a fixed point
  not on the line called the focus. Will be a U
  shaped graph.
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• Projectile an object shot through the air, such
  as a football, that has Independent Vertical and
  Horizontal motions and after receiving an initial
  thrust travels through the air only under the
  force of gravity. From Old Book, motion of
  objects given an initial velocity that then move
  only under the force of gravity.
• Trajectory the path of a projectile through
  space.

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INDEPENDENCE OF MOTION IN TWO DIMENSIONS

• Example of dropping a softball and launching one
  horizontally at 2 m/s. In both cases the
  horizontal acceleration is ZERO. (Dropped ball
  does not move horizontally and launched ball had
  a constant velocity thus no acceleration. (See
  Figure 6-1)
• Also in this example you see that the dropped and
  launched balls have the same vertical motion.
  Both balls are accelerated downward by the force
  of gravity. Both balls would hit the ground at
  the same time. Similar to the boat taking the
  same time to get across the river if there were
  no flow downstream and if there was a current
  downstream.
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• The horizontal motion of the thrown ball does not
  affect its vertical motion at all. The
  horizontal and vertical components are
  Independent of each other.

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INDEPENDENCE OF MOTION IN TWO DIMENSIONS

• The combination of a Constant Horizontal Velocity
  and Uniform Vertical Acceleration (Gravity)
  produces a Trajectory that has a Parabolic Shape.
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• Since Horizontal and Vertical parts are
  Independent of each other if you find the time of
  one the other is the same, similar to the boat
  across the river problems.
• The shape of the trajectory and the horizontal
  motion depend on the viewpoint or frame of
  reference of the observer, But the Vertical
  Motion does not.

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INDEPENDENCE OF MOTION IN TWO DIMENSIONS

• x vxt where x is the horizontal displacement,
  vx is the initial horizontal velocity, and t is
  the time
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• vxf vi where vxf is the final horizontal
  velocity and vi is the initial velocity
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• y vyt ½ gt2 where y is the vertical
  displacement, vy is the initial vertical
  velocity, t is time, and g is gravity
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• vyf vy gt where vyf is the final vertical
  velocity and vy is initial vertical velocity
• Do Practice Problems p. 150 1-3

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PROJECTILES LAUNCHED AT AN ANGLE

• When a projectile is launched at an angle the
  initial velocity has a vertical and horizontal
  component.
• Max Height the height of the projectile when
  the vertical velocity is zero. The max height
  occurs when the time is HALF of the entire flight
  time (Example if time is 5 seconds then the time
  for the max height is 2.5 seconds).
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• Range denoted by R the horizontal distance
  between the launch point of the projectile and
  where it returns to launch height or the
  horizontal distance from the point of bounce
  until the projectile returns to the surface
  height.
•  
• Range is the horizontal distance traveled during
  the entire flight time.

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PROJECTILES LAUNCHED AT AN ANGLE

• vx vi cos ? or Ax Ai cos ?
• vy vi sin ? or Ay Ai sin ?

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PROJECTILES LAUNCHED AT AN ANGLE

• Do Example 1 p. 151
• vx vi cos ? 4.5(cos 66) 4.5(.407) 1.83
  m/s
• vy vi sin ? 4.5(sin 66) 4.5(.914) 4.11
  m/s
• A) Time b) Max Height
• y vyt ½ gt2 Since the trajectory is
  symmetric
• 0 vyt ½ gt2 the max height occurred at
  .4165 s
• -4.11t ½ (-9.8)t2 So y vyt ½ gt2
• t t y 4.11(.4195) ½
  (-9.8)(.4195)2
• -4.11 -4.9t y 1.724 (-4.9)(.176)
• -4.11 / -4.9 t y 1.724 - .862
• .839 s t y .862 m
• C) On next slide 

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PROJECTILES LAUNCHED AT AN ANGLE

• c) Range is the horizontal distance traveled
  during the entire flight time.
• So R vxt
• R 1.83(.839)
• R 1.535 m
• Do Practice Problems p. 152 4-6

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TRAJECTORIES DEPEND UPON THE VIEWER

• Remember that the force due to air resistance
  exists and it can be important but for now we are
  ignoring it.
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• Do 6.1 Section Review p. 152 7-11