chapter 2_Projectile_Motion final (1) (1).pdf

physics

1. CHAPTER 2
Projectile Motion
Joufuniversity
Faculty of Science and Arts in
Qurayyat.
Physics Department
PHS 211
Projectile Motion
Prepared by Physics Department

2. What is projectile?
Projectile -Any object which projected by some
means and continues to move due to its own
inertia (mass).

3. Projectiles move in TWO dimensions
Since a projectile moves in 2-dimensions, it
therefore has 2 components just like a
resultant vector.

Horizontal

Motion of a ball rolling freely along a
level surface

Horizontal velocity is ALWAYS constant
Vertical

Vertical

Motion of a freely falling object

Force due to gravity

Vertical component of velocity changes
with time

Parabolic

Path traced by an object accelerating
only in the vertical direction while moving
at constant horizontal velocity

4. Summary: Equations of liner motion &
free-fall

X-Component

Y-Component
tvxx
xiif 
gttvyy 
1
2

Vectors
gtvv
ygvv
gttvyy
yiyf
yiyf
yiif



2
2
1
22
2
)sin(
)cos(


iyi
ixi
vv
vv


Note: g= 9.8
m/s^2

5. Horizontal “Velocity” Component

NEVER changes, covers equal displacements in equal
time periods. This means the initial horizontal velocity
equals the final horizontal velocity
constant
ox x
v v 
Projectiles which have NO upward
trajectory and NO initial VERTICAL
velocity.
In other words, the horizontal velocity is
CONSTANT.
BUT WHY?
Gravity DOES NOT work horizontally to
increase or decrease the velocity.
0 /
oy
v m s

6. Equations for horizontally launched projectiles
Horizontal Motion
Δx = V
x
Δt
V
x
= V
xi
= constant
Vertical Motion
Δy = +½ g •(Δt)
2
Δy = +½ g •(Δt)
2
V
yf
= +g •Δt
V
yf
2
= +2g •Δy
Overall Final Velocity
V
f
2
= V
x
2
+ V
yf
2

7. Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for
the “y” direction. And for this we use kinematic #2.
21
2
ox
x vt at 
2
ox
x vt
Remember, the velocity is
CONSTANT horizontally, so
that means the acceleration
is ZERO!
21
2
y gt
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.

8. Vertically Launched Projectiles
There are several things you must consider when doing these types
of projectiles besides using components. If it begins and ends
at ground level, the “y” displacement is ZERO: y = 0

Changes (due to gravity), does NOT cover equal displacements
in equal time periods.
Both the MAGNITUDE and DIRECTION
change. As the projectile moves up the
MAGNITUDE DECREASES and its
direction is UPWARD. As it moves down
the MAGNITUDE INCREASES and the
direction is DOWNWARD.

9. Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
v
o
v
oy
ox
x vt
21
2
oy
y vt gt 
v
o
v
ox
v
oy

ox
2
oy
cos
sin
ox o
oy o
v v
v v





10. Combining the Components
Together, these components produce what is called a trajectory or
path. This path is parabolicin nature.
Vertical
Velocity
decreases
Vertical Velocity
increases on
the way down,
NO Vertical Velocity at the top of the trajectory.
ComponentMagnitudeDirection
HorizontalConstantConstant
VerticalChangesChanges
Horizontal
Velocity is
constant
decreases
on the way
upward
the way down,

11. Horizontally Launched Projectiles
Example:
Aplanetravelingwith
ahorizontalvelocityof100
m/sis500mabovethe
ground.Atsomepointthe
pilotdropsabombona
targetbelow.(a)Howlong
What do I
know?
What I want to
know?
v
ox
=100 m/s
t = ?
y = 500 m
x = ?
v
oy
= 0 m/s
targetbelow.(a)Howlong
isthebombintheair?(b)
Howfarawayfrompoint
abovewhereitwasdropped
willitland?
g = -9.8 m/s/s
2 2
2
1 1
500 ( 9.8)
2 2
102.04
y gt t
t t
   
  
10.1 seconds
(100)(10.1)
ox
x v t  
1010 m

12. Projectiles Equations
Since the projectile was launched at a angle, the velocity MUST
be broken into components!!!
v
o
v
ox
v
oy

Initial Velocity
Angle

17. Example 1:
A place kicker kicks a football with a velocity of 20.0
m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
  53
cos
20cos53 12.04 /
sin
20sin53 15.97 /
ox o
ox
oy o
oy
v v
v m s
v v
v m s



 

 

18. Example2:
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(a) How long is the ball in
the air?
What I knowWhat I want
to know
v
ox
=12.04 m/s
t = ?
v
oy
=15.97 m/s
x = ?
y = 0 y
max
=?
g = -9.8
m/s/sm/s/s
2 2
2
1 0 (15.97) 4.9
2
15.97 4.9 15.97 4.9
oy
y v t gt t t
t t t
t
    
   

3.26 s

19. Follow Example 2
A
place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(b) How far away does it
What I knowWhat I want
to know
v
ox
=12.04 m/s
t = 3.26 s
v
oy
=15.97 m/s
x = ?
y = 0 y=?
(b) How far away does it
land?
y = 0 y
max
=?
g = -9.8
m/s/s
(12.04)(3.26)
ox
x vt  
39.24 m

20. Follow Example 2
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(c) How high does it
What I knowWhat I want to
knowt = 3.26 s
x = 39.24 m
y = 0 y
max
=?
(c) How high does it
travel?
CUT YOUR TIME IN HALF!
g = -9.8
m/s/s
2
2
1
2
(15.97)(1.63) 4.9(1.63)
oy
y v t gt
y
y
 
 

13.01 m

21. Example 3
A place kicker kicks a
football with a velocity
of 10.0 m/s and at an
angle of 30 degrees.
(a)
Find the maximum
high
(b)
Rang?
What I knowWhat I want to
know
Max high h
max
=?
Rang (R)=m
y = 0
g = -9.8 m/s/s
(b)
Rang?
(c)
Total time of travel ?
1.02 sec


22. Example 4
An object is launched from the
base of an incline, which is at
an angle of 30°. If the launch
angle is 60°from the
horizontaland the launch
speed is 10 m/s, what is the
total flight time?
What I knowWhat I
want to
know
T=?
θ= 60
g = -10 m/s2
total flight time?
g = -10 m/s2

Solution: In order to account for the incline angle, we have to
reorient the coordinate system so that the points of projection
and return are on the same level. The angle of projection with
respect to thexdirection isθ−α, and the acceleration in
theydirection isgcosα. We replaceθwithθ−αandgwithg cosα:

23. 1.15 sec


24. o
0 m/s 40 m/s sin53
3.26 s
9.8 m/s
y yo y
y yo
y
v v a t
v v
t
a
 
  
  

Example:
How long does it take to reach
maximum height, y
max
?
At maximum height, v
y
= 0 m/s
24
 
 
 
m .
m/s .
sin53m/s /sm
m
a
vv
yy
yyavv
2
o22
y
yoy
o
oyyoy
152
892
400
0
2
2
222
22







9.8 m/s
y
a 
What is the maximum height?

25. When is the projectile
at y= 25m?
 
1
2
1
2
 tatvyy
yyoo
25
 

s 61.5 and s 910.0
2
s 1.54)s 52.6()s 52.6(
s 0s 10.5s 52.6
0
22
0
2
1
22
222
2
2








t
tt
a
yy
t
a
v
t
yytvta
y
o
y
yo
oyoy

26. What are the velocity
components then, at
t = 0.910 s and t = 5.61 s?
tm/s m/s
tm/s m/s tgsinvtavv
m/s m/s vtavv
2
2o
ooyyoy
o
ooxxox



8.99.31
)8.9(53sin40)(
1.2453cos40cos


26
time (s) velocity (m/s)
0.910
5.61
0.23
1.24


y
x
v
v
023
124
.v
.v
y
x


tm/s m/s
2
 8.99.31

27. Example:
How far does the object travel in the x-direction?
21
tatvxx 
We need to know the elapsed time, t.
27
2
2
1
tatvxx
xoxo 
We need to know the elapsed time, t.
The total elapsed time is the time it takes to go up plus the time it takes to come down.
Previously, we found that the time to reach maximum height was t= 3.26 s.
The total time, then, is 2x3.26s = 6.52 s. [Verify with .]

ms.
s
m
.tvx
tvtatvxx
ox
oxxoxo
157526124
00
2
1
2


2
89
2
1
00 t.ty
max 

28. What are the velocity & position components at t= 3 seconds?
Example:
28


m 1.44s3m/s 8.9
2
1
3sm/s 0m 0
2
1
m60s3m/s 20m 0
m/s 4.293s)m/s 8.9(m/s 0
m/s 203m/s 0m/s 20
222
2
2




tatvyy
tvxx
tavv
stavv
yyoo
xoo
yyoy
xxox
What are the velocity & position components at t= 3 seconds?

29. Class Exercise
An object is fired from the ground at 100
meters per second at an angle of 30
degrees with the horizontal

Calculate the horizontal and vertical

Calculate the horizontal and vertical
components of the initial velocity

After 2.0 seconds, how far has the object
traveled in the horizontal direction?

How high is the object at this point?

30. Solution

Part a

Part b
 
 

s
m
s
m
vv
s
m
s
m
vv
iiy
iix
5030sin100sin
8730cos100cos
0
0





Part b

Part c


ms
s
m
tvx
t
x
v
x
ix
1740.287 




 

 

2
2
2
0.28.9
2
1
0.250
2
1
s
s
m
s
s
m
tgtvy
iy
