Projectile Motion or 2D Kinematics

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Projectile Motion or 2D Kinematics

• References
• Conceptual Physics, Paul G. Hewitt, 10th edition,
  Addison Wesley publisher
• http//www.glenbrook.k12.il.us/gbssci/Phys/Class/v
  ectors/u3l2a.html

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Outline

• What is a projectile
• Characteristics of a projectile's motion
• Horizontal and vertical components of velocity
  and displacement
• Initial velocity components
• Examples of problems

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What is a projectile?

• A projectile is an object upon which the only
  force acting is gravity. A projectile is any
  object which once projected or dropped continues
  in motion by its own inertia and is influenced
  only by the downward force of gravity.

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Projectile Motion and Inertia

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Horizontal and Vertical Velocities

• A projectile is any object upon which the only
  force is gravity,
• Projectiles travel with a parabolic trajectory
  due to the influence of gravity,
• There are no horizontal forces acting upon
  projectiles and thus no horizontal acceleration,
• The horizontal velocity of a projectile is
  constant (a never changing value),
• There is a vertical acceleration caused by
  gravity its value is 9.8 m/s/s, down,
• The vertical velocity of a projectile changes by
  9.8 m/s each second,
• The horizontal motion of a projectile is
  independent of its vertical motion.

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Vector diagrams for projectile motion
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Horizontal and vertical displacement
Horizontally Launched Projectile
If the horizontal displacement (x) of a
projectile were represented by an equation, then
that equation would be written as x vix t
y 0.5 g t2 (equation for vertical
displacement for a horizontally launched
projectile)?
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Displacement diagram of projectile motion

• Horizontal

Vertical
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Horizontal and vertical displacement Non -
Horizontally Launched Projectile
y viy t 0.5 g t2 (equation for vertical
displacement for an angled-launched projectile)?
If the horizontal displacement (x) of a
projectile were represented by an equation, then
that equation would be written as x vix t
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Check your understanding
1. Anna Litical drops a ball from rest from the
top of 78.4-meter high cliff. How much time will
it take for the ball to reach the ground and at
what height will the ball be after each second of
motion? 2. A cannonball is launched horizontally
from the top of an 78.4-meter high cliff. How
much time will it take for the ball to reach the
ground and at what height will the ball be after
each second of travel?
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Answers

• 1.
• It will take 4 seconds to fall 78.4 meters
• Use the equation y 0.5 g t2 and substitute
  -9.8 m/s/s for g. The vertical displacement must
  then be subtracted from the initial height of 78.
  4 m.
• At t 1 s, y 4.9 m (down) so height is 73.5 m
  (78.4 m - 4.9 m )?
• At t 2 s, y 19.6 m (down) so height is 58.8 m
  (78.4 m - 19.6 m )?
• At t 3 s, y 44.1 m (down) so height is 34.3 m
  (78.4 m - 45 m)?
• At t 4 s, y 78.4 m (down) so height is 0 m
  (78.4 m - 78.4 m)?
• 2.
• It will take 4 seconds to fall 78.4 meters
• Use the equation y 0.5 g t2 and substitute
  -9.8 m/s/s for g. The vertical displacement must
  then be subtracted from the initial height of 78.
  4 m.
• At t 1 s, y 4.9 m (down) so height is 73.5 m
  (78.4 m - 4.9 m )?
• At t 2 s, y 19.6 m (down) so height is 58.8 m
  (78.4 m - 19.6 m )?
• At t 3 s, y 44.1 m (down) so height is 34.3 m
  (78.4 m - 45 m)?
• At t 4 s, y 78.4 m (down) so height is 0 m
  (78.4 m - 78.4 m)?

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Check your understanding
3. Fill in the table indicating the value of the
horizontal and vertical components of velocity
and acceleration for a projectile. 4. The diagram
below shows the trajectory for a projectile
launched non-horizontally from an elevated
position on top of a cliff. The initial
horizontal and vertical components of the
velocity are 8 m/s and 19.6 m/s respectively.
Positions of the object at 1-second intervals are
shown. Determine the horizontal and vertical
velocities at each instant shown in the diagram.
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Answers

• 3.The vx values will remain constant at 15.0 m/s
  for the entire 6 seconds the ax values will be 0
  m/s/s for the entire 6 seconds.
• The vy values will be changing by -9.8 m/s each
  second. Thus,
• vy 29.4 m/s (t 0 s) vy 19.6 m/s (t 1
  s)?
• vy 9.8 m/s (t 2 s) vy 0 m/s (t 3 s)?
• vy -9.8 m/s (t 4 s) vy -19.6 m/s (t 5
  s)?
• vy -29.4 m/s (t 6 s)?
• The ay values will be -9.8 m/s/s for the entire 6
  seconds.
• 4.The vx values will remain 8 m/s for the entire
  6 seconds.
• The vy values will be changing by 9.8 m/s each
  second. Thus,
• vy 9.8 m/s (t 1 s) vy 0 m/s (t 2 s)?
• vy -9.8 m/s (t 3 s) vy -19.6 m/s (t 4
  s)?
• vy -29.4 m/s (t 5 s) vy -39.2 m/s (t 6
  s)?

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Initial components of velocity
vx v cos a vy v sin a
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Evaluating various info

• Determination of the Time of Flight
• Determination of Horizontal Displacement
• x vix t
• Determination of the Peak Height
• y viy t 0.5 g t2

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Equations of motion
the vertical acceleration of a projectile is
known to be -9.8 m/s/s

• Horizontal motion
• Vertical Motion

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Solving Projectile Motion Problems

• The following procedure summarizes the above
  problem-solving approach.
• Carefully read the problem and list known and
  unknown information in terms of the symbols of
  the kinematic equations. For convenience sake,
  make a table with horizontal information on one
  side and vertical information on the other side.
• Identify the unknown quantity which the problem
  requests you to solve for.
• Select either a horizontal or vertical equation
  to solve for the time of flight of the
  projectile.
• With the time determined, use one of the other
  equations to solve for the unknown. (Usually, if
  a horizontal equation is used to solve for time,
  then a vertical equation can be used to solve for
  the final unknown quantity.)

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Check your understanding

• A football is kicked with an initial velocity of
  25 m/s at an angle of 45-degrees with the
  horizontal. Determine the time of flight, the
  horizontal displacement, and the peak height of
  the football.

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Answer

• State the problem
• Use the appropriate equations of motion

• The unknown quantities are the horizontal
  displacement, the time of flight, and the height
  of the football at its peak.
• From the vertical information in the table above
  and the second equation listed among the vertical
  kinematic equations (vfy viy ayt), it
  becomes obvious that the time of flight of the
  projectile can be determined. By substitution of
  known values, the equation takes the form
  of -17.7 m/s 17.7 m/s (-9.8 m/s/s)t
• -35.4 m/s (-9.8 m/s/s)t
• 3.61 s t
• The total time of flight of the football is 3.61
  seconds.
• With the time determined, information in the
  table and the horizontal kinematic equations can
  be used to determine the horizontal displacement
  (x) of the projectile. The first equation (x
  vixt 0.5axt2) listed among the horizontal
  kinematic equations is suitable for determining
  x. With the equation selected, the physics
  problem once more becomes transformed into an
  algebra problem. By substitution of known values,
  the equation takes the form of x (17.7
  m/s)(3.6077 s) 0.5(0 m/s/s)(3.6077 s)2
• x (17.7 m/s)(3.6077 s)?
• x 63.8 m
• The horizontal displacement of the projectile is
  63.8 m.
• Finally, the problem statement asks for the
  height of the projectile at is peak. This is the
  same as asking, "what is the vertical
  displacement (y) of the projectile when it is
  halfway through its trajectory?" In other words,
  find y when t 1.80 seconds (one-half of the
  total time). To determine the peak height of the
  projectile (y with t 1.80 sec), the first
  equation (y viyt 0.5ayt2) listed among the
  vertical kinematic equations can be used. By
  substitution of known values into this equation,
  it takes the form of
• y (17.7 m/s)(1.80 s) 0.5(-10 m/s/s)(1.80
  s)2
• y 31.9 m (-15.9 m)?
• y 15.9 m
• The solution to the problem statement yields the
  following answers the time of flight of the
  football is 3.61 s, the horizontal displacement
  of the football is 63.8 m, and the peak height of
  the football 15.9 m.

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The Problem-Solving Approach

• The following procedure summarizes the above
  problem-solving approach.
• Use the given values of the initial velocity (the
  magnitude and the angle) to determine the
  horizontal and vertical components of the
  velocity (vix and viy).
• Carefully read the problem and list known and
  unknown information in terms of the symbols of
  the kinematic equations. For convenience sake,
  make a table with horizontal information on one
  side and vertical information on the other side.
• Identify the unknown quantity which the problem
  requests you to solve for.
• Select either a horizontal or vertical equation
  to solve for the time of flight of the
  projectile. For non-horizontally launched
  projectiles, the second equation listed among the
  vertical equations (vfy viy ayt) is usually
  the most useful equation.
• With the time determined, use a horizontal
  equation (usually x vixt 0.5axt2 ) to
  determine the horizontal displacement of the
  projectile.
• Finally, the peak height of the projectile can be
  found using a time value which one-half the total
  time of flight. The most useful equation for this
  is usually y viyt 0.5ayt2 .

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SUMMARY See Hand-outs