# 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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