Projectile Motion A. Projectile Motion B. Horizontal Projectile Motion C. Projectile Motion

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Projectile MotionA. Projectile
MotionB. Horizontal Projectile MotionC.
Projectile Motion At an AngleD. Projectile
Motion Time of FlightE. Projectile Motion
Practice Problems
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A. Projectile Motion

• Projectile
• Any object upon which the only force acting is
  gravity
• Examples
• An object dropped at rest (air resistance is
  negligible)
• An object throw upwards at an angle (air
  resistance is negligible)

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A. Projectile Motion

• Free body (Force) Diagram for a projectile
• Remember that forces are only required to cause
  an acceleration (not a motion)

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A. Projectile Motion
Horizontal Motion Vertical Motion
Forces present No Yes Fgrav acts downward
Acceleration No Yes g9.81 m/s2
Velocity Constant Changing 10 m/s
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A. Projectile Motion

• Two Types
• Launched Horizontally
• Launched at an angle

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B. Horizontal Projectile Motion

• A cannonball is shot horizontally from a very
  high cliff at a high speed
• In the absence of gravity, what would the motion
  of the cannonball be like?
• Continue in motion in a straight line at constant
  speed

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B. Horizontal Projectile Motion

• In the presence of gravity
• Parabolic trajectory
• Gravity causes a vertical acceleration

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B. Horizontal Projectile Motion

• NOTE
• Due to the absence of horizontal forces, a
  projectile remains in motion with a constant
  horizontal velocity
• Horizontal forces are not required to keep a
  projectile moving horizontally
• The only force acting upon a projectile is
  gravity!

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B. Horizontal Projectile Motion

• A cannonball is launched horizontally off a high
  cliff with an initial speed of 20 m/s
• Vertical velocity is changing 10 m/s every
  second
• Horizontal velocity remains constant

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B. Horizontal Projectile Motion
Time (s) Horizontal Velocity Vertical Velocity
0 s 20 m/s, right 0 m/s
1 s 20 m/s, right 10 m/s, down
2 s 20 m/s, right 20 m/s, down
3 s 20 m/s, right 30 m/s, down
4 s 20 m/s, right 40 m/s, down
5 s 20 m/s, right 50 m/s, down
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B. Horizontal Projectile Motion

• Projectile released from rest with no horizontal
  velocity
• Horizontal and Vertical Displacement
• Horizontal
  Displacement
• x vix t
• Vertical
  Displacement
• y 0.5 g t2

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B. Horizontal Projectile Motion
Time (s) Horizontal Displacement Vertical Displacement
0 s 0 m 0 m
1 s 25 m 5 m
2 s 50 m 20 m
3 s 75 m 45 m
4 s 100 m 80 m
5 s 125 m 125 m
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C. Projectile Motion At an Angle

• Projectile launched at an angle to the horizontal
• Horizontal and Vertical displacement

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C. Projectile Motion At an Angle

• Two Components
• Horizontal motion
• Vertical motion

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C. Projectile Motion At an Angle

• If the cannonball is launched at an angle, how do
  the horizontal and vertical velocity values
  change?

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C. Projectile Motion At an Angle
Time (s) Horizontal Velocity Vertical Velocity
0 s 34.6 m/s, right 20 m/s, up
1 s 34.6 m/s, right 10 m/s, up
2 s 34.6 m/s, right 0 m/s
3 s 34.6 m/s, right 10 m/s, down
4 s 34.6 m/s, right 20 m/s, down
5 s 34.6 m/s, right 30 m/s, down
6 s 34.6 m/s, right 40 m/s, down
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C. Projectile Motion At an Angle

• Since the cannonball was not launched
  horizontally but at an angle, the equation to
  find the vertical component for velocity and
  displacement is as follows
• y viy t 0.5 g t2

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C. Projectile Motion At an Angle

• Calculate the vertical and horizontal components
  for t 1 s
• y viy t 0.5 g t2
• where viy 20 m/s
• y (20 m/s)(1 s) 0.5(-10m/s2)(1 s) 2
• y 20 m (-5 m)
• y 15 m
• x vix t
• where vix 34.6 m/s
• x (34.6 m/s) (1 s)x 34.6 m

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C. Projectile Motion At an Angle

• Complete the following chart in your notes

Time (s) Horizontal Displacement Vertical Displacement
0 s 0 m 0 m
1 s 34.6 m 15 m
2 s
3 s
4 s
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C. Projectile Motion At an Angle

• Complete the following chart in your notes

Time (s) Horizontal Displacement Vertical Displacement
0 s 0 m 0 m
1 s 34.6 m 15 m
2 s 69.2 m 20 m
3 s 103.8 m 15 m
4 s 138. 4 m 0 m
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C. Projectile Motion At an Angle

• Determining the initial velocity of a projectile
• Horizontal and vertical component
• Use trig functions and the angle at which the
  projectile is launched

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C. Projectile Motion At an Angle

• A projectile is launched with an initial velocity
  of 50 m/s at an angle of 60 above the
  horizontal.
• Determine the vertical and horizontal components
  of the initial velocity

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C. Projectile Motion At an Angle

• cos 60 vx / 50 m/s
• vx 50 m/s (cos 60)
• vx 25 m/s
• sin 60 vy / 50 m/s
• vy 50 m/s (sin 60)
• vy 43 m/s

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C. Projectile Motion At an Angle

• A water balloon is launched with a speed of 40
  m/s at an angle of 60 to the horizontal
• Determine the horizontal and vertical components
  of the initial velocity
• vx 20 m/s
• vy 34.6 m/s

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D. Projectile Motion Time of Flight

• Determination of the Time of Flight
• The time for a projectile to rise vertically to
  its peak (as well as the time to fall from the
  peak) is dependent upon the vertical motion

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D. Projectile Motion Time of Flight

• Time for a projectile to reach its peak
• tup viy / g
• Once you have determined the time it takes for a
  projectile to reach its peak, now you can find
  the total time
• t total 2 (tup)

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D. Projectile Motion Time of Flight
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D. Summary Projectile Motion
Kinematics Equations Horizontal Components Vertical Components
d vi t .5 at2 x vixt.5 ax t2 y viy t.5 ay t2
Vf vi a t Vfx vix ax t Vfy viy ay t
Vf2 vi2 2a d Vfx2vix22a ax d Vfy2viy22ay d
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E. Projectile Motion Practice Problems

• A soccer ball is kicked horizontally off a
  22.0-meter high hill and lands a distance of 35.0
  meters from the edge of the hill. Determine the
  initial horizontal velocity of the soccer ball.
• 16.7 m/s

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E. Projectile Motion Practice Problems

• A long jumper leaves the ground with an initial
  velocity of 12 m/s at an angle of 28-degrees
  above the horizontal. Determine the time of
  flight, the horizontal distance, and the peak
  height of the long-jumper.
• t 1.1 s
• 11.9 m
• 1.6 m