Warm-Up 09/13/10

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Warm-Up 09/13/10

• Please express the Graphic Vector Addition Sums
  in MAGNITUDE-ANGLE format (last two pages of
  PhyzJob packet)

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Projectile Motion

• Projectile Motion motion in two dimensions
  (horizontal and vertical) with the vertical
  motion under the action of gravity only
  (downward).
• The initial velocity is in the horizontal
  direction.
• Because the action of gravity is in the vertical
  direction, the horizontal motion has zero
  acceleration if air resistance is ignored.
• The vertical motion is a free fall, so the
  acceleration due to gravity, g, is -9.80 m/s2

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Because the only acceleration is that of gravity,
the time it takes for the projectile to reach the
ground is the same as the time it would take if
the object were simply dropped.
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Equations for Projectile Motion Resolve the
motion into x and y components
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Example 1 A package is dropped from an airplane
traveling with a constant horizontal speed of 120
m/s at an altitude of 500 m. What horizontal
distance, or range, does the package travel
before hitting the ground?
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Solution for Example 1
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Problem-Solving Tip

• The quantities such as initial velocities and
  displacements have to be treated independently.
• The initial horizontal velocity is 120 m/s and
  the initial vertical velocity is 0 m/s. 120 m/s
  can only be used in the horizontal motion and the
  0 m/s can be used only in the vertical motion.
• A common mistake is to use the 120 m/s for the
  vertical motion.

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Example 2 A golfer hits a golf ball with a
velocity of 35 m/s at an angle of 25 o above the
horizontal. If the point where the ball is hit
and the point where the ball lands are at the
same level, A) How much time does the ball spend
in the air?B) What horizontal distance does the
ball travel before landing?
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Setting Up Example 2

• Given
• horizontal motion vertical motion
• X00 m y0 0 m
• Vx0 V0cosq Vy0V0sinq
• y 0 m
• Vx0 (35 m/s) cos 25o Vy0 (35m/s)sin25o
• 31.7 m/s 14.8 m/s
• Find A) t and B) x
• yy0 Vy0t ½ gt2, solve quadratic for t
• x x0 Vx0t

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SolutionA) The t0 corresponds to the starting
position and the t 3.02 s corresponds to the
landing position, so the time of flight is 3.0 s
( 2 sig figs)B)
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Regardless of its path, a projectile will always
follow these rules

1. Projectiles always maintain a constant horizontal
   velocity (neglecting air resistance)
2. Projectiles always experience a constant vertical
   acceleration of 9.8 m/s2 downward (neglecting air
   resistance)
3. Horizontal and vertical motion are completely
   independent of one another. Therefore, the
   velocity of a projectile can be separated into
   horizontal and vertical components.

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Cont.

1. For a projectile beginning and ending at the same
   height, the time it takes to rise to its highest
   point equals the time it takes to fall from the
   peak back to the original height.
2. Objects dropped from a moving vehicle have the
   same velocity as the moving vehicle
3. In order to solve projectile exercises, you MUST
   consider horizontal and vertical motion
   separately.

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• Concept Test slides 14-41
• D\Chapter_03\Assess\Assess_Present\WBL6_ConcepTes
  ts_Ch03.ppt
• Physlet Exploration 3.5
• Homework p 98 100 61, 66, 67, 70, 75, 79, 81,
  83, 84, BONUS 71

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Warm-Up 09/15/09

• A small plane takes off at a constant velocity of
  150 km/h at an angle of 37. At 3.00 s, (a) how
  high is the plane above the ground, and (b) what
  horizontal distance has the plane traveled from
  the liftoff point?  

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Warm-Up 09/21/09

• A ball is thrown horizontally out the window of
  a building with a velocity of 6.5 m/s from a
  height of 100m. How far from the base of the
  building will the ball land? Tip find time, t,
  using the equation yy0 Vy0t ½(gt2), then
  use t to calculate x by
• x x0 Vx0t