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PROJECTILE MOTION
WAYNE FERNANDES
Introduction
 Projectile Motion:
Motion through the air without a propulsion
 Examples:
Part 1.
Motion of Objects Projected
Horizontally
v0
x
y
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
constant
•The horizontal and
vertical motions are
independent of each other,
but they have a common
time
g = -9.81m/s2
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?
x
y
0
Frame of reference:
h
v0
Equations of motion:
X
Uniform m.
Y
Accel. m.
ACCL. ax = 0 ay = g = -9.81
m/s2
VELC. vx = v0 vy = g t
DSPL. x = v0 t y = h + ½ g t2
g
Trajectory
x = v0 t
y = h + ½ g t2
Eliminate time, t
t = x/v0
y = h + ½ g (x/v0)2
y = h + ½ (g/v0
2
) x2
y = ½ (g/v0
2
) x2
+ h
y
x
h
Parabola, open down
v01
v02 > v01
Total Time, Δt
y = h + ½ g t2
final y = 0 y
x
h
ti =0
tf =Δt
0 = h + ½ g (Δt)2
Solve for Δt:
Δt = √ 2h/(-g)
Δt = √ 2h/(9.81ms-2
)
Total time of motion depends
only on the initial height, h
Δt = tf - ti
Horizontal Range, Δx
final y = 0, time is
the total time Δt
y
x
h
Δt = √ 2h/(-g)
Δx = v0 √ 2h/(-g)
Horizontal range depends on the
initial height, h, and the initial
velocity, v0
Δx
x = v0 t
Δx = v0 Δt
VELOCITY
v
vx = v0
vy = g t
v = √vx
2
+ vy
2
= √v0
2
+g2
t2
tg Θ = v / v = g t / v
Θ
FINAL VELOCITY
v
vx = v0
vy = g t
v = √vx
2
+ vy
2
v = √v0
2
+g2
(2h /(-g))
v = √ v0
2
+ 2h(-g)
Θ
tg Θ = g Δt / v0
= -(-g)√2h/(-g) / v0
= -√2h(-g) / v0
Θ is negative
(below the
horizontal line)
Δt = √ 2h/(-g)
HORIZONTAL THROW - Summary
Trajectory Half -parabola, open
down
Total time Δt = √ 2h/(-g)
Horizontal Range Δx = v0 √ 2h/(-g)
Final Velocity v = √ v0
2
+ 2h(-g)
tg Θ = -√2h(-g) / v0
h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2
Part 2.
Motion of objects projected at an
angle
vi
x
y
θ
vix
viy
Initial velocity: vi = vi [Θ]
Velocity components:
x- direction : vix = vi cos Θ
y- direction : viy = vi sin Θ
Initial position: x = 0, y = 0
x
y
• 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
a = g =
- 9.81m/s2
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
Uniform motion
Y
Accelerated motion
ACCELERATION ax = 0 ay = g = -9.81 m/s2
VELOCITY vx = vix= vi cos Θ
vx = vi cos Θ
vy = viy+ g t
vy = vi sin Θ + g t
DISPLACEMENT x = vix t = vi t cos Θ
x = vi t cos Θ
y = h + viy t + ½ g t2
y = vi tsin Θ + ½ g t2
Equations of motion:
X
Uniform motion
Y
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 tsin Θ + ½ g t2
Trajectory
x = vi t cos Θ
y = vi tsin Θ + ½ g t2
Eliminate time, t
t = x/(vi cos Θ)
y
x
Parabola, open down
2
22
22
2
cos2
tan
cos2cos
sin
x
v
g
xy
v
gx
v
xv
y
i
ii
i
Θ
+Θ=
Θ
+
Θ
Θ
=
y = bx + ax2
Total Time, Δt
final height y = 0, after time interval Δt
0 = vi Δtsin Θ + ½ g (Δt)2
Solve for Δt:
y = vi tsin Θ + ½ g t2
0 = vi sin Θ + ½ g Δt
Δt =
2 vi sin Θ
(-g)
t = 0 Δt
x
Horizontal Range, Δx
final y = 0, time is
the total time Δt
x = vi t cos Θ
Δx = vi Δt cos Θ
x
Δx
y
0
Δt =
2 vi sin Θ
(-g)
Δx =
2vi
2
sin Θ cos Θ
(-g)
Δx =
vi
2
sin (2 Θ)
(-g)
sin (2 Θ) = 2 sin Θ cos Θ
Horizontal Range, Δx
Δx =
vi
2
sin (2 Θ)
(-g)
Θ (deg) sin (2Θ)
0 0.00
15 0.50
30 0.87
45 1.00
60 0.87
75 0.50
90 0
•CONCLUSIONS:
•Horizontal range is greatest for the
throw angle of 450
• Horizontal ranges are the same for
angles Θ and (900
– Θ)
Trajectory and horizontal range
2
22
cos2
tan x
v
g
xy
i Θ
+Θ=
0
5
10
15
20
25
30
35
0 20 40 60 80
15 deg
30 deg
45 deg
60 deg
75 deg
vi = 25 m/s
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 tsin Θ + ½ g t2
At maximum height vy = 0
0 = vi sin Θ + g tup
tup =
vi sin Θ
(-g)
tup = Δt/2
hmax = vi tupsin Θ + ½ g tup
2
hmax = vi
2
sin2
Θ/(-g) + ½ g(vi
2
sin2
Θ)/g2
hmax =
vi
2
sin2
Θ
2(-g)
Projectile Motion – Final Equations
Trajectory Parabola, open down
Total time Δt =
Horizontal range Δx =
Max height hmax =
(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2
2 vi sin Θ
(-g)
vi
2
sin (2 Θ)
(-g)
vi
2
sin2
Θ
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
THANK YOU

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Projectile motion

  • 2. Introduction  Projectile Motion: Motion through the air without a propulsion  Examples:
  • 3. Part 1. Motion of Objects Projected Horizontally
  • 5. x y
  • 6. x y
  • 7. x y
  • 8. x y
  • 9. x y •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 g = -9.81m/s2
  • 10. 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?
  • 11. x y 0 Frame of reference: h v0 Equations of motion: X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = -9.81 m/s2 VELC. vx = v0 vy = g t DSPL. x = v0 t y = h + ½ g t2 g
  • 12. Trajectory x = v0 t y = h + ½ g t2 Eliminate time, t t = x/v0 y = h + ½ g (x/v0)2 y = h + ½ (g/v0 2 ) x2 y = ½ (g/v0 2 ) x2 + h y x h Parabola, open down v01 v02 > v01
  • 13. Total Time, Δt y = h + ½ g t2 final y = 0 y x h ti =0 tf =Δt 0 = h + ½ g (Δt)2 Solve for Δt: Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms-2 ) Total time of motion depends only on the initial height, h Δt = tf - ti
  • 14. Horizontal Range, Δx final y = 0, time is the total time Δt y x h Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v0 Δx x = v0 t Δx = v0 Δt
  • 15. VELOCITY v vx = v0 vy = g t v = √vx 2 + vy 2 = √v0 2 +g2 t2 tg Θ = v / v = g t / v Θ
  • 16. FINAL VELOCITY v vx = v0 vy = g t v = √vx 2 + vy 2 v = √v0 2 +g2 (2h /(-g)) v = √ v0 2 + 2h(-g) Θ tg Θ = g Δt / v0 = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 Θ is negative (below the horizontal line) Δt = √ 2h/(-g)
  • 17. HORIZONTAL THROW - Summary Trajectory Half -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v0 √ 2h/(-g) Final Velocity v = √ v0 2 + 2h(-g) tg Θ = -√2h(-g) / v0 h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2
  • 18. Part 2. Motion of objects projected at an angle
  • 19. vi x y θ vix viy Initial velocity: vi = vi [Θ] Velocity components: x- direction : vix = vi cos Θ y- direction : viy = vi sin Θ Initial position: x = 0, y = 0
  • 20. x y • 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 a = g = - 9.81m/s2
  • 21. 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?
  • 22. Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.81 m/s2 VELOCITY vx = vix= vi cos Θ vx = vi cos Θ vy = viy+ g t vy = vi sin Θ + g t DISPLACEMENT x = vix t = vi t cos Θ x = vi t cos Θ y = h + viy t + ½ g t2 y = vi tsin Θ + ½ g t2
  • 23. Equations of motion: X Uniform motion Y 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 tsin Θ + ½ g t2
  • 24. Trajectory x = vi t cos Θ y = vi tsin Θ + ½ g t2 Eliminate time, t t = x/(vi cos Θ) y x Parabola, open down 2 22 22 2 cos2 tan cos2cos sin x v g xy v gx v xv y i ii i Θ +Θ= Θ + Θ Θ = y = bx + ax2
  • 25. Total Time, Δt final height y = 0, after time interval Δt 0 = vi Δtsin Θ + ½ g (Δt)2 Solve for Δt: y = vi tsin Θ + ½ g t2 0 = vi sin Θ + ½ g Δt Δt = 2 vi sin Θ (-g) t = 0 Δt x
  • 26. Horizontal Range, Δx final y = 0, time is the total time Δt x = vi t cos Θ Δx = vi Δt cos Θ x Δx y 0 Δt = 2 vi sin Θ (-g) Δx = 2vi 2 sin Θ cos Θ (-g) Δx = vi 2 sin (2 Θ) (-g) sin (2 Θ) = 2 sin Θ cos Θ
  • 27. Horizontal Range, Δx Δx = vi 2 sin (2 Θ) (-g) Θ (deg) sin (2Θ) 0 0.00 15 0.50 30 0.87 45 1.00 60 0.87 75 0.50 90 0 •CONCLUSIONS: •Horizontal range is greatest for the throw angle of 450 • Horizontal ranges are the same for angles Θ and (900 – Θ)
  • 28. Trajectory and horizontal range 2 22 cos2 tan x v g xy i Θ +Θ= 0 5 10 15 20 25 30 35 0 20 40 60 80 15 deg 30 deg 45 deg 60 deg 75 deg vi = 25 m/s
  • 29. Velocity •Final speed = initial speed (conservation of energy) •Impact angle = - launch angle (symmetry of parabola)
  • 30. Maximum Height vy = vi sin Θ + g t y = vi tsin Θ + ½ g t2 At maximum height vy = 0 0 = vi sin Θ + g tup tup = vi sin Θ (-g) tup = Δt/2 hmax = vi tupsin Θ + ½ g tup 2 hmax = vi 2 sin2 Θ/(-g) + ½ g(vi 2 sin2 Θ)/g2 hmax = vi 2 sin2 Θ 2(-g)
  • 31. Projectile Motion – Final Equations Trajectory Parabola, open down Total time Δt = Horizontal range Δx = Max height hmax = (0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) vi 2 sin2 Θ 2(-g)
  • 32. PROJECTILE MOTION - SUMMARY  Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration  The projectile moves along a parabola