4.1 Independence of
horizontal and vertical
motions
• Monkey and hunter experiment
• Body projected horizontally
under gravity
• Body projected at an angle q
under gravity
1
4.1 Independence of horizontal and vertical motions (SB p. 144)
What is “projectile motion”?
Kicking a football
The football will be projected
along ____________ path
2
4.1 Independence of horizontal and vertical motions (SB p. 144)
Independence of horizontal and
vertical motions
Assume air resistance is negligible. During the
flight,
(1)the horizontal velocity ______________
_______________
(2)
the vertical velocity is subjected to a
_________________________
3
4
4.1 Independence of horizontal and vertical motions (SB p. 145)
Monkey and hunter experiment
2
breaks the aluminium
strip and the circuit of the
electromagnet
3
1 bullet leaves the
mouth of the toy
gun
iron monkey
falls vertically
downwards
4 the bullet
always hits
the monkey
5
4.1 Independence of horizontal and vertical motions (SB p. 145)
Monkey and hunter experiment
Why does the bullet always hit the monkey?
The bullet and the monkey start with
_______ vertical downward velocity
They fall under the same acceleration due to gravity
their vertical motions are __________
Their _______ positions are
the same at any instant of time
6
4.1 Independence of horizontal and vertical motions (SB p. 145)
Monkey and hunter experiment
We can conclude that:
The horizontal and vertical components
of a projectile motion are ____________
of each other.
7
4.1 Independence of horizontal and vertical motions (SB p. 146)
Body projected horizontally under gravity
a marble is projected horizontally
off the edge of a table
travels with a uniform velocity
in the horizontal direction and
accelerates uniformly
downwards due to gravity
the horizontal velocity
remains constant
8
4.1 Independence of horizontal and vertical motions (SB p. 146)
Body projected horizontally under gravity
horizontal motion
vertical
motion
They are independent
of each other
9
10
4.1 Independence of horizontal and vertical motions (SB p. 152)
Body projected at an angle q under gravity
Initial vertical
velocity:
uy = u sinq
the object is projected with a velocity
u at an angle q to the horizontal
Initial horizontal velocity:
ux = _________
11
4.1 Independence of horizontal and vertical motions (SB p. 152)
Body projected at an angle q under gravity
•
The vertical motion of the object undergoes
constant acceleration a = –g
•
Vertical component of velocity (vy) and the
vertical displacement (y) at time t :
vy 

y 

12
4.1 Independence of horizontal and vertical motions (SB p. 152)
Body projected at an angle q under gravity
•
The horizontal motion is uniform and the object
moves with constant velocity
•
Horizontal velocity component (vx) and the
horizontal displacement (x) at time t:
vx 

x

13
4.1 Independence of horizontal and vertical motions (SB p. 152)
Body projected at an angle q under gravity
•
Instantaneous velocity (v):
v
• The direction of motion at certain instant:
tanφ 
14
4.1 Independence of horizontal and vertical motions (SB p. 153)
Body projected at an angle q under gravity
At B, the object reaches its maximum height H
with vy = ___,
v y 2  uy 2  2as

 H
15
4.1 Independence of horizontal and vertical motions (SB p. 153)
Body projected at an angle q under gravity
At D, the time of flight of the object (T)
θ
 T  2usin
g
16
4.1 Independence of horizontal and vertical motions (SB p. 153)
Body projected at an angle q under gravity
Horizontal displacement OD of the projectile is
the range R
R

 R
2
u sin2θ
g
17
4.1 Independence of horizontal and vertical motions (SB p. 153)
Body projected at an angle q under gravity
By combining:
x
and y 
 y  x tan θ -







g
2
2
2u cos θ







x2
where
0o < q < 90o
The equation is of the parabolic form y = ax – bx2
trajectory of the projectile is a __________
18
4.1 Independence of horizontal and vertical motions (SB p. 154)
Body projected at an angle q under gravity
Note:
1.The vertical and horizontal motions of a projectile
motion are independent of each other
2. The trajectory of the projectile is ___________
3. The maximum value of R is _______
occurs when sin2q = _______
q = ____

maximum range is obtained if the object is
projected at an angle ____ to the horizontal
19
4.1 Independence of horizontal and vertical motions (SB p. 154)
Body projected at an angle q under gravity
Note:
4.  sin2q = sin(180o – 2q)
two angles of projection for a given range R
with a specific speed, except for q = 45o
Maximum range at q = 45o
Range at 60o =
Range at 30o
20
A projectile is launched with an initial velocity of 30
m s-1 at an angle of 60o above the horizontal.
Calculate the magnitude and direction of its velocity 5
s after launch. (Take g = 9.8 m s-2)
Ans : 27.5 m s-1, 56.9o
below the horizontal
21
A police officer is chasing you across a roof top, both
are running at 4.5 m s-1. Before you reaches the edge
of the roof, you has to decide whether or not to try
jumping to the roof of the next building, which is 6.2
m away but 4.8 m lower. Can you make it?
22
Effect of Atmosphere
In an projectile experiment, the launch angle is 60 and
the launch speed is 100 mi/h and the following results
are obtained :
Path I (Vacuum)
Path II (Air)
Range
581 ft
323 ft
Maximum height
252 ft
174 ft
Time of flight
7.9 s
6.6 s
Effects of Air Resistance :
(i)
(ii)
(iii)
(iv)
The trajectory becomes asymmetric.
23
4.1 Independence of horizontal and vertical motions (SB p. 154)
Body projected at an angle q under gravity
Note:
5. In real situation, air resistance reduces the speed
and the range of the projectile
Go to
Go to
More to Know 1
Example 3
24
End
25
4.1 Independence of horizontal and vertical motions (SB p. 148)
Q: An aeroplane, flying in a straight line at a
constant height of 500 m with a speed of
200 m s–1, drops an object. The object takes
a time t to reach the ground and travels a
horizontal distance d in doing so. Taking g as
10 m s–2 and ignoring air resistance, what are
the values of t and d ?
Solution
26
4.1 Independence of horizontal and vertical motions (SB p. 148)
Solution:
When the object is released from
the aeroplane,
Horizontal component of velocity
= Velocity of the aeroplane
= 200 m s–1
For the vertical motion of the object:
Initial velocity = 0,
acceleration = g = 10 m s–2,
displacement = 500 m
1
Using s  ut  at 2
500 = 0 +
t2
= 100
t = 10 s
1
2
2
 10  t2
Return to
For the horizontal
motion:
Distance (d) = 200  t
= 200  10
= 2 000 m
Text
27
4.1 Independence of horizontal and vertical motions (SB p. 149)
Q: When a rifle is fired horizontally at a target P on a
screen at a range of 25 m, the bullet strikes the
screen at a point 5.0 mm below P. The screen is now
moved to a distance of 50 m from the rifle and the
rifle again is fired horizontally at P in its new position.
Assuming that air resistance may be neglected,
what is the new distance below P at which the
screen would now be struck?
Solution
28
4.1 Independence of horizontal and vertical motions (SB p. 149)
Solution:
Let v = velocity of the bullet when it leaves the rifle.
 Time taken for the bullet to travel through a horizontal distance of 25 m
(t1) =
25
v
Consider vertical motion of the bullet,
initial velocity (u) = 0, acceleration (a) = g, displacement (s) = 5.0  10–3 m,
time (t1) =
Using
25
v
1
2
s  ut  at
2
2
1  25 
3
50  10  0  g   1
2  v 
29
4.1 Independence of horizontal and vertical motions (SB p. 149)
Solution (cont’d):
With the screen at 50 m from the rifle, time of travel (t2) = 50
v
Using
1
s  ut  at
2
2
2
1  50 
New distance below P (h) = 0  g    2 
2  v 
(2)  (1)
2
 50 

4
-3  25 
5.0  10
h

h  20.0  10
 20 mm
-3
Return to
Text
30
4.1 Independence of horizontal and vertical motions (SB p. 154)
So far, air resistance has been
neglected for projectile motions. In real
situations, the trajectory of the
projectile is influenced by air resistance
in the extent that depends on the mass,
shape and size of the objects.
Return to
Text
31
4.1 Independence of horizontal and vertical motions (SB p. 154)
Q: An aeroplane flies at a height h with a
constant horizontal velocity u so as to fly
over a cannon. When the aeroplane is
directly over the cannon, a shell was fired to
hit the aeroplane. Neglecting air resistance,
what is the minimum speed of the shell in
order to hit the aeroplane?
(g = acceleration due to gravity)
Solution
32
4.1 Independence of horizontal and vertical motions (SB p. 154)
Solution:
Let
v = minimum speed of shell,
q = angle of projection.
The minimum speed is when the shell will hit the aeroplane at the maximum
height of the shell trajectory. Since the shell is fired when the aeroplane is
directly above it, the horizontal component of velocity of the shell must be
the same as the velocity of the aeroplane, i.e. same u.
33
4.1 Independence of horizontal and vertical motions (SB p. 154)
Solution (cont’d):
Maximum height 
Using the equation,

h
2
2
v sin θ
2g
2
2
v sin θ
2g
2
2
v sin θ  2 gh  (1)
Horizontal component of velocity of the shell,
v cos θ  u
v2cos 2θ  u2(2)
(1)  (2)

v2sin 2θ  v2cos 2θ  u2  2gh

v  u2  2gh
Return to
Text
34
4.2 Terminal velocity
(Vertical motion under
gravity with air resistance)
35
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air
falling under influence of air
resistance
net force acting on him is
due to his weight (mg)
and the upward air
resistance (f)
By applying Newton’s
Second Law:
ma = mg – f
36
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air

his velocity (v) increases
the air resistance acting on him becomes
larger
f = bv
where
b = constant
37
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air
•
moves faster and faster
the air resistance
= his weight
net force acting on him = 0
reaches the
terminal
velocity (vTA)
with 0
acceleration
38
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air
reaches
terminal
velocity (vTA)
with 0
acceleration
0 = mg - bvTA
mg
v

b
T
A
39
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air
air resistance > parachutist’s weight
resultant upward net
force acting on him
slows him down
air resistance
40
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 157)
A parachutist falling in the air
the air
resistance
will be
balanced by
the weight
again
at a new lower terminal velocity vTC
before he lands safely
41
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 158)
A parachutist falling in the air
42
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
Independence of horizontal
and vertical motions
1. Assume air resistance is negligible, in a
projectile motion, the horizontal velocity
remains constant, and the vertical velocity
is subjected to a constant acceleration due
to gravity.
2. The horizontal velocity (vx) and vertical
velocity (vy) are independent of each other
in projectile motions.
43
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
Independence of horizontal
and vertical motions
3. For objects projected horizontally under
gravity:
(a) The instantaneous velocity (v) of the
projectile is:
v v
x
2
v
y
2
44
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
Independence of horizontal
and vertical motions
(b) The direction of
motion is
determined by:
(c) If the projectile is
projected at a
height H above the
ground, its time of
flight (tf) is:
tanφ 
v
v
y
x
2H
t 
g
f
45
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
Independence of horizontal
and vertical motions
(d) The total horizontal
displacement travelled
by the object is called
the range (R):
(e) The trajectory of the
projectile is a
parabola:
2H
R u
x g
where ux is the
initial horizontal
velocity


 g  2
y  -
x
 2u 2 
 x 
46
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
Independence of horizontal
and vertical motions
4. For objects projected with initial
velocity (u) at an angle q under
gravity:
(a) Its maximum height (H) is:
u 2 sin2θ
H
2g
47
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.1
(b)
Independence of horizontal
and vertical motions
Its time of flight (tf) is:
t 
f
2u sinθ
g
q
R  u2 sin2
g
(c)
Its range (R) is:
(d)
The trajectory of the
projectile is a

 2
g

x
y

x
tan
θ
 2

parabola:
2
 2u cos θ
48
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.2
Terminal velocity
5. Consider a parachutist of mass m falling under
the influence of air resistance. As the
parachutist falls, its velocity (v) increases. When
the velocity increases, the air resistance acted
on the parachutist also increases. We have:
f = bv
where f is the air resistance and b is a constant.
49
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 159)
4.2
Terminal velocity
6. The terminal velocity of the parachutist (vT) is:
mg
v 
b
T
where g is the acceleration due
to gravity.
50
4.2 Terminal velocity (Vertical motion under gravity with air resistance (SB p. 160)
51
End
52