PROJECTILE MOTION 12 by elipldoc

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									             PROJECTILE MOTION
            Senior High School Physics
                         Lech Jedral
                            2006
  Part 1.                              Part 2.

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Introduction
           Motion:
 Projectile
Motion through the air without a propulsion
 Examples:
          Part 1.
Motion of Objects Projected
       Horizontally
y   v0




         x
y




    x
y




    x
y




    x
y




    x
y

                    •Motion is accelerated
                    •Acceleration is constant,
                    and downward
                    • a = g = -9.81m/s2
                    •The horizontal (x)
                    component of velocity is
    g = -9.81m/s2   constant
                    •The horizontal and vertical
                    motions are independent of
                    each other, but they have a
                    common time          x
ANALYSIS OF MOTION
ASSUMPTIONS:
•    x-direction (horizontal):       uniform motion
•    y-direction (vertical):         accelerated motion
•    no air resistance
QUESTIONS:
•    What is the trajectory?
•    What is the total time of the motion?
•    What is the horizontal range?
•    What is the final velocity?
 Frame of reference:       Equations of motion:

        y
                                   X           Y
                v0
                              Uniform m.   Accel. m.
                        ACCL.   ax = 0   ay = g = -9.81
h           g                                 m/s2
                        VELC.       vx = v 0         vy = g t

                       x DSPL.      x = v0 t      y = h + ½ g t2
    0
                          Trajectory
     x = v0 t                     y
     y = h + ½ g t2                   Parabola, open down
                              h
      Eliminate time, t
t = x/v0
y = h + ½ g (x/v0)2                     v01        v02 > v01

y = h + ½ (g/v02) x2
y = ½ (g/v02) x2 + h
                                                      x
                    Total Time, Δt          Δt = tf - ti
y = h + ½ g t2
final y = 0                         y

0 = h + ½ g (Δt)2                   ti =0
                                h
Solve for Δt:

 Δt = √ 2h/(-g)

Δt = √ 2h/(9.81ms-2)
                                              tf =Δt
Total time of motion depends
only on the initial height, h                       x
              Horizontal Range, Δx
   x = v0 t
  final y = 0, time is               y
  the total time Δt

    Δx = v0 Δt                   h


 Δt = √ 2h/(-g)

Δx = v0 √ 2h/(-g)
                                         Δx
Horizontal range depends on the
initial height, h, and the initial
                                              x
velocity, v0
                           VELOCITY

                                     vx = v0

                                       Θ
                          vy = g t
                                               v
v = √vx  2   + vy   2


= √v02+g2t2

tg Θ = v / v = g t / v
        y x           0
                      FINAL VELOCITY

                                   vx = v0

  Δt = √ 2h/(-g)                     Θ           tg Θ = g Δt / v0
                        vy = g t
                                             v   = -(-g)√2h/(-g) / v0
v = √vx2   + vy   2
                                                 = -√2h(-g) / v0
v = √v02+g2(2h /(-g))
                                                  Θ is negative
v=√   v02+   2h(-g)                               (below the
                                                  horizontal line)
HORIZONTAL THROW - Summary
h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2

  Trajectory                        Half -parabola, open
                                    down
  Total time                        Δt = √ 2h/(-g)

  Horizontal Range                  Δx = v0 √ 2h/(-g)

  Final Velocity                    v = √ v02+ 2h(-g)
                                    tg Θ = -√2h(-g) / v0
            Part 2.
Motion of objects projected at an
             angle
  y




          vi   Initial position: x = 0, y = 0

               Initial velocity: vi = vi [Θ]
viy
               Velocity components:
               x- direction : vix = vi cos Θ
               y- direction : viy = vi sin Θ
      θ
                                                x
      vix
   y

a =g=
- 9.81m/s2
             • Motion is accelerated
             • Acceleration is constant, and
               downward
             •   a = g = -9.81m/s2
             • The horizontal (x) component of
               velocity is constant
             • The horizontal and vertical
               motions are independent of each
               other, but they have a common
               time


                                                 x
ANALYSIS OF MOTION:
ASSUMPTIONS
•   x-direction (horizontal):       uniform motion
•   y-direction (vertical):         accelerated motion
•   no air resistance
QUESTIONS
•   What is the trajectory?
•   What is the total time of the motion?
•   What is the horizontal range?
•   What is the maximum height?
•   What is the final velocity?
               Equations of motion:

                        X                         Y
                  Uniform motion        Accelerated motion
ACCELERATION    ax = 0                 ay = g = -9.81 m/s2

VELOCITY        vx = vix= vi cos Θ     vy = viy+ g t
                vx = vi cos Θ          vy = vi sin Θ + g t
DISPLACEMENT    x = vix t = vi t cos Θ y = h + viy t + ½ g t2
                x = vi t cos Θ         y = vi t sin Θ + ½ g t2
               Equations of motion:

                        X                        Y
                  Uniform motion       Accelerated motion
ACCELERATION    ax = 0                ay = g = -9.81 m/s2

VELOCITY        vx = vi cos Θ         vy = vi sin Θ + g t


DISPLACEMENT    x = vi t cos Θ        y = vi t sin Θ + ½ g t2
                          Trajectory
x = vi t cos Θ
                                       Parabola, open down
y = vi t sin Θ + ½ g t2           y
Eliminate time, t
       t = x/(vi cos Θ)
   vi x sin      gx 2
y             2
   vi cos 2vi cos2 
                      g
y  x tan                 x2
                2vi2 cos2 

  y = bx + ax2
                                                    x
                      Total Time, Δt
y = vi t sin Θ + ½ g t2
final height y = 0, after time interval Δt
0 = vi Δt sin Θ + ½ g (Δt)2
  Solve for Δt:

  0 = vi sin Θ + ½ g Δt                             x


         2 vi sin Θ
  Δt =                                       t=0   Δt
          (-g)
            Horizontal Range, Δx
  x = vi t cos Θ                                  y
  final y = 0, time is
  the total time Δt

 Δx = vi Δt cos Θ
        2 vi sin Θ
 Δt =                                                                  x
          (-g)                                   0
                     sin (2 Θ) = 2 sin Θ cos Θ
                                                               Δx
       2vi 2 sin Θ cos Θ                              vi 2 sin (2 Θ)
Δx =                                      Δx =
            (-g)                                        (-g)
            Horizontal Range, Δx
                      vi 2 sin (2 Θ)
               Δx =
                        (-g)

Θ (deg) sin (2 Θ)
                       •CONCLUSIONS:
0       0.00
                       •Horizontal range is greatest for the
15      0.50           throw angle of 450
30      0.87
45      1.00           • Horizontal ranges are the same for
60      0.87           angles Θ and (900 – Θ)
75      0.50

90      0
         Trajectory and horizontal range
                                g
              y  x tan   2          x2
                           2vi cos 2 

35

30
                       vi = 25 m/s
                                            15 deg
25                                          30 deg
                                            45 deg
20
                                            60 deg
15
                                            75 deg
10

5

0
     0       20           40           60            80
                      Velocity




•Final speed = initial speed (conservation of energy)
•Impact angle = - launch angle (symmetry of parabola)
                Maximum Height
vy = vi sin Θ + g t
y = vi t sin Θ + ½ g t2
  At maximum height vy = 0

0 = vi sin Θ + g tup      hmax = vi t upsin Θ + ½ g tup2
                          hmax = vi2 sin2 Θ/(-g) + ½ g(vi2 sin2 Θ)/g2
        vi sin Θ
tup =
         (-g)                        vi2 sin2 Θ
                          hmax =
                                     2(-g)
tup = Δt/2
         Projectile Motion – Final Equations
(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2
    Trajectory            Parabola, open down




                                   2 vi sin Θ
    Total time            Δt =
                                    (-g)


                                   vi 2 sin (2 Θ)
    Horizontal range      Δx =
                                       (-g)

                                     vi2 sin2 Θ
    Max height            hmax =
                                      2(-g)
 PROJECTILE MOTION - SUMMARY
 Projectile motion is motion with a constant
  horizontal velocity combined with a constant
  vertical acceleration
 The projectile moves along a parabola

								
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