Projectile Motion. Determining the Range
Let us examine the motion of a projectile in X and Y
and then determine the Range (R), the distance travelled
in the case the projectile lands at the same elevation as it
was released.
y  y f  yi  0
R
R  x f  xi
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Page 2.96
The only thing common between horizontal and vertical
motion is the time taken. So we will use it to relate these
two motions.
vix  vi cos 
viy  vi sin 
In x:
vi

x f  xi  vix t
 R  x f  xi  vi cos( )t
Solve for t:
R
t
vi cos( )
APSC 111 Kinematics
Page 2.97
vix  vi cos 
viy  vi sin 
In y:
vi

y f  yi  viy t  at
1
2
2
y  y f  yi  0
0  viy t  12 at 2
0  vi sin( )t  12 at 2
vi sin( )   12 at
Solve for t:
2vi sin( )
t
a
APSC 111 Kinematics
Page 2.98
From x:
R
t
vi cos( )
Solve for R:
From y:
2vi sin( )
t
a
2vi sin( )
R
vi cos( )
a
2sin( ) cos( )  sin(2 )
vi2 sin(2 )
R  
a
v sin(2 )
R 
g
2
i
APSC 111 Kinematics
a  9.81   g
(Only when Δy=0)
Page 2.99
v sin(2 )
R
g
2
i
Note: Maximum range occurs when
sin(2 )  1
2  90
0
  450
(Only when Δy=0)
And, as usual, we are ignoring the effects of air
resistance, which can be substantial !
APSC 111 Kinematics
Page 2.100
Uniform Circular Motion
Uniform circular motion: motion in a circle of
constant radius at constant speed
v
r
Instantaneous velocity: is always tangent to the
circle.
APSC 111 Kinematics
Page 2.101
Uniform Circular Motion
Constant speed: what can we say about acceleration?
v
v
r
Because the velocity is changing in direction, the
object is accelerating !
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Page 2.102
Uniform Circular Motion
There is a centripetal, or radial, acceleration which as we
shall see, points toward the center of the circle.

a1
We shall also show that
v2
a
r

a2
another case of constant
acceleration….
APSC 111 Kinematics
Page 2.103
Analysis of uniform circular motion:
Some facts:
vx
vx  v sin 
v y  v cos 
v
x  r cos 
y  r sin 
r

Position of object is:

vy
y
x

r  xiˆ  yjˆ
and velocity is:

v  vx iˆ  v y ˆj
APSC 111 Kinematics
Page 2.104

Which can be written as: v  v sin  iˆ  v cos  ˆj
or, with x  r cos  , y  r sin 
vx
vy ˆ vx ˆ

v  i
j
r
r
v
y
v
Differentiate to get acceleration:


v dy ˆ v dx ˆ
 dv
a
i

j
dt
r dt
r dt
y

v ˆ v ˆ

a   v y i  vx j
x
r
r
r
2
v
v

2
2
( v y )   vx  
a a
r
r
APSC 111 Kinematics
as claimed
Page 2.105
What about angle?
v ˆ v ˆ

a   v y i  vx j
r
r
2
v2
v

a   cos  iˆ  sin  ˆj
r
r
Note: components give a in –x and –y
directions.
tan  
ay
ax
  

v2
 sin 
r
v2
 cos
r
 tan 
vx
v

a
ax
vy
ay
a points to centre
APSC 111 Kinematics
Page 2.106
Example Problem:
20cm
wall
floor
The disk is rotating with a period of 5.0 ms. When the red blob stuck
to the disk is at “5 o’clock”, it comes free. If it is at a height of 1.2 m
at this point, how high up the wall will it hit. The wall is 2.5 m away.
APSC 111 Kinematics
Page 2.107
Example Problem:
20cm
wall
floor
Start by calculating initial velocity and angle:
2 r 2 (0.2)

 251.3 m/s
T  5.00ms  v 
T
.005
APSC 111 Kinematics



1
360  30
12
Page 2.108
20cm
wall
Velocity components:
vix  vi cos   217.7 m/s
viy  vi sin   125.7 m/s
Time in flight:
x f  xi  vix t
2.5 m
t
 0.011 s
217.7 m/s
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Page 2.109
20cm
wall
In this time:
y f  yi  viy t  12 at 2
y f  1.20  125.7(0.011)  (9.81)(0.011)
1
2
2
y f  2.58 m
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Page 2.110
Relative Velocity
We have already considered relative speed in one
dimension:
5 km/h (relative to train)
80 km/h
85 km/h (relative to ground)
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Page 2.111
Relative Velocity
Each velocity is labeled first with the object, and second with the
reference frame in which it has this velocity. So for the train
example, we would write:
vMG  vMT  vTG
Where M is the Man, T is the Train, and G is the Ground
vMG  5  80  85 km/h
Note: If the man accelerates in the train: then the acceleration as
observed from the ground is:
aMG
d
  vMT  vTG   aMT
dt
The acceleration is the same in both frames (Train and Ground)
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Page 2.112
Relative Velocity in 2 Dimensions
It is similar in two dimensions except that we must add and
subtract velocities as vectors.



vMG  vMT  vTG
For example if the man was moving at some angle across the train.
Again, the acceleration is independent of the frame if the man
(object) accelerates:
d 



aMG   vMT  vTG   aMT
dt
APSC 111 Kinematics
Page 2.113
Example: A boat moving across a river with
a perpendicular current
Here, vWS is the velocity of the
water in the shore frame, vBS is
the velocity of the boat in the
shore frame, and vBW is the
velocity of the boat in the water
frame.
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Page 2.114
Relative Velocity
Example: Heading upstream.
A boat’s speed in still water
is vBW = 1.85 m/s. If the boat
is to travel directly across a
river whose current has
speed vWS = 1.20 m/s, at
what upstream angle must
the boat head?



vBS  vBW  vWS
APSC 111 Kinematics
Page 2.115
Relative Velocity
Example: Heading upstream.
1.20
sin  
1.85
1.2
  40
1.85
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Page 2.116
Example: Heading across the river.
The same boat (vBW = 1.85 m/s) now
heads directly across the river whose
current is still 1.20 m/s. (a) What is the
velocity (magnitude and direction) of the
boat relative to the shore? (b) If the river
is 110 m wide, how long will it take to
cross and (c) how far downstream will the
boat be then?



vBS  vBW  vWS
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Page 2.117
• What is the velocity (magnitude and
direction) of the boat relative to the
shore?
1.20
tan  
1.85
1.2
1.85
  33
vBS  (1.2) 2  (1.85) 2  2.2 m/s
APSC 111 Kinematics
Page 2.118
(b) If the river is 110 m wide, how long
will it take to cross and how far
downstream will the boat be then?
tan(33) = y / 110
y = 71 m
110 m
330
Vx = x / t → t = x/Vx = 110/1.85
t = 59.5 s
APSC 111 Kinematics
Page 2.119
Relative Velocity
• The compass of an airplane indicates that it is headed
due north, and its airspeed indicator shows that it is
moving through the air at 240 km/h. If there is a west
wind of 100 km/h, what it the velocity of the plane
relative to the



 ground?
v AG  100 km h
vPG  vPA  v AG
vPG  240  100  260 km/h
2

vPA  240 km h
N
2
100
tan  
240

  22.6 East of North

APSC 111 Kinematics
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Relative Velocity
- In which direction should the pilot head to travel due north?
- What will then be her velocity relative to the ground?

v AG  100 km h



vPG  vPA  v AG
N
v 2 PG  1002  2402
vPG  218 km/h

vPA  240 km h
100
sin  
240

  24.6 West of North
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Page 2.121