PROJECTILE MOTION
Senior High School Physics
Lech Jedral
2006
Part 1.
Part 2.
Introduction
 Projectile
Motion:
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
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
x
common time
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:
y
v0
g
h
0
Equations of motion:
X
Y
Uniform m.
Accel. m.
ACCL.
ax = 0
ay = g = -9.81
m/s2
VELC.
vx = v 0
vy = g t
x DSPL.
x = v0 t
y = h + ½ g t2
Trajectory
y
x = v0 t
y = h + ½ g t2
Eliminate time, t
t = x/v0
y = h + ½ g (x/v0)2
Parabola, open down
h
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
y
final y = 0
0 = h + ½ g (Δt)2
Solve for Δt:
h
ti =0
Δt = √ 2h/(-g)
Δt = √ 2h/(9.81ms-2)
Total time of motion depends
only on the initial height, h
tf =Δt
x
Horizontal Range, Δx
x = v0 t
y
final y = 0, time is
the total time Δt
Δ x = v0 Δ t
h
Δt = √ 2h/(-g)
Δx = v0 √ 2h/(-g)
Horizontal range depends on the
initial height, h, and the initial
velocity, v0
Δx
x
VELOCITY
vx = v0
Θ
v = √vx
2
+ vy
2
= √v02+g2t2
tg Θ = v / v = g t / v
y x
0
vy = g t
v
FINAL VELOCITY
vx = v0
Δt = √ 2h/(-g)
v = √vx
2
+ vy
2
vy = g t
v = √v02+g2(2h /(-g))
v=√
v02+
2h(-g)
tg Θ = g Δt / v0
Θ
v
= -(-g)√2h/(-g) / v0
= -√2h(-g) / v0
Θ is negative
(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
θ
y- direction : viy = vi sin 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
Uniform motion
ax = 0
Y
Accelerated motion
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 y = h + viy t + ½ g t2
x = vi t cos Θ
y = vi t sin Θ + ½ g t2
ACCELERATION
Equations of motion:
X
Uniform motion
ax = 0
Y
Accelerated motion
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
ACCELERATION
Trajectory
x = vi t cos Θ
y = vi t sin Θ + ½ g t2
Parabola, open down
y
Eliminate time, t
t = x/(vi cos Θ)
y
v i x sin 
v i cos 
y  x tan  

gx
2
2 v i cos 
2
2
g
2 v i cos 
2
2
x
2
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
Δt =
x
2 vi sin Θ
(-g)
t=0
Δt
Horizontal Range, Δx
x = vi t cos Θ
y
final y = 0, time is
the total time Δt
Δx = vi Δt cos Θ
Δt =
2 vi sin Θ
(-g)
Δx =
x
sin (2 Θ) = 2 sin Θ cos Θ
2vi 2 sin Θ cos Θ
(-g)
0
Δx =
Δx
vi 2 sin (2 Θ)
(-g)
Horizontal Range, Δx
Δx =
Θ (deg) sin (2 Θ)
0
0.00
15
0.50
30
0.87
45
1.00
60
0.87
75
0.50
90
0
vi 2 sin (2 Θ)
(-g)
•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
y  x tan  
g
2 v i cos 
2
2
x
2
35
vi = 25 m/s
30
15 deg
30 deg
25
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
tup =
vi sin Θ
(-g)
tup = Δt/2
hmax = vi t upsin Θ + ½ g tup2
hmax = vi2 sin2 Θ/(-g) + ½ g(vi2 sin2 Θ)/g2
hmax =
vi2 sin2 Θ
2(-g)
Projectile Motion – Final Equations
(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2
Trajectory
Total time
Horizontal range
Parabola, open down
Δt =
Δx =
2 vi sin Θ
(-g)
vi 2 sin (2 Θ)
(-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