Position and Displacement
If we want to locate a point in space we can use a position vector
that extends from a reference point to the location we are trying
to describe.




r  xi  y j  zk
i, j, k are the unit vectors
along the x, y, z, directions
x, y, z are the components
of the vector along those
directions (i, j, k)
the displacement is just
the change in a particle’s position




r  3 i  2 j  5 k
If we have an initial and final positions described by two position
vectors, ri and rf




ri  xi i  yi j  zi k




rf  x f i  y f j  z f k
So our displacement is defined as:
  
 r  rf  ri



 x f  xi i   y f  yi  j  z f  zi k
Let’s look at an example:




ri  3 m i  2m j  5m k




rf  9m i  2m j  8m k




ri  3m i  2m j  5m k




rf  9m i  2m j  8m k
  
 r  rf  ri



 9m   3m  i  2m  2m  j  8m  5m  k


 12 m i  3m k
the definitions for velocity and acceleration are basically
the same but you have to remember that these are vectors
quantities:
average velocity:

v Avg
  
 r rf  ri


t
t
referring to the previous example, if it took the particle
3 seconds to to the displacement, then the average velocity
is:

v Avg



 r 12 i  3 k
m m 


 6 i 1 k
t
3
s
s
similarly, the instantaneous velocity is:

 dr
v
dt
GO TO HITT
acceleration follows the same rrr-guments:

a Avg
 v  v
v
f
i


t
t
instantaneous acceleration is:

 dv
a
dt
the acceleration vector shows the
direction of the acceleration of the
particle
Motion in more than one dimension.
http://www.physicsclassroom.com/mmedia/vectors/mzi.html
straight line
(line of site)
the monkey begins to fall ad the
precise moment when the ball
leaves the barrel of the gun
The ball and the
monkey arrive at
the point marked
by the red
dot at the
same
time
path of
bullet
path of monkey
Two types of motion occurring !
Monkey: Object under constant acceleration in
1 – dimension
y
Bullet: The motion of the bullet is a combination
x
of motion with a constant velocity (along
the x-axis) and motion with constant
acceleration (along the y-axis)
To start analyzing this problem we must find the x, and y components
of the velocity:
vi
y



vi  vixi  viy j
viy opp

vix adj
x
opp viy
sin  

 viy  vi sin 
hyp vi
adj vix
cos  

 vix  vi cos 
hyp vi
Let’ start by looking at an example where the launch angle is zero
h=1m
v0X = 2 m/s
=0
x=?
x  v0 x t
h  v0 y t 
1 2
gt
2
1 2 1 2
v0 y  0  h  v0 y t  g t  g t
2
2
1 2
h  g t solve for time, t
2
2h
2h
2
2
2h  g t  t 
 t
g
g
now that we have an expression for time, t we can find the range x
h=1m
v0X = 2 m/s
=0
x  v0 x t
x=?
vf
We could also find the
final velocity, vf
t
2h
g
x  v0 x
2h
m 21m
2
 0.9 m
m
g
s 9.8
s2



2
2
v fy  v0 y  gt
v f  v fx x  v fy y
v f  v fx   v fy 
m


2
2
2
2
v f  v fx   v fy   0   9.8 2 0.45s 
m
m

 

s 

  2    4.5 
s
 s 
m
m
 4.5
v fx  v0 x  2
m
s
s
 4.8
s
v fy  ??
Let’s work an example that is a launch angle different from 0,
but it returns to the same height that it is launched.
A pirate ship is 560 m from a fort defending the harbor
entrance of an island. A defense cannon, located at sea
level, fires balls at initial speed v0 = 82 m/s
At what angle,  from the horizontal must a ball be fired to hit
the ship?
x  v0 x t
 v0 cos  t
We need an expression for the time, t
x  v0 x t
 v0 cos  t
We need an expression for the time, t
1 2
h  y  y0  v0 y t  gt
2
since the projectile will
return to the same height that it left from,
h = y –y0 = 0 so our equation becomes:
1 2
1 2
0  v0 y t  gt  v0 sin  t  gt
2
2
1 2
0  v0 sin  t  gt
2
from here we can solve for time, t
factor out a t, and it cancels:
1
1
0  v0 sin   gt  gt  v0 sin 
2
2
2v0 sin 
gt  2v0 sin   t 
g
now substitute this back into:
x  v0 cos  t
 2v0 sin   2v02
 
x  v0 cos  t  v0 cos  
cos  sin 
g
g


from trigonometry:
v02
x  sin 2
g
sin 2  2sin  cos
this is called the Range equation!
DANGER WILL ROBINSON: This equation only works if the projectile
returns to the same height that it was launched!!!
v02
gx 9.8m / s 2 560m
x  sin 2  sin 2  2 
g
v0
82m / s 2
 0.816
2  sin 1 0.816
 54.6 or 180  54.6  125.3
  27 o or 63o
Now let’s look at an example of projectile motion where the
projectile lands at a different elevation from it’s launch height.
A stone is projected at a cliff of height h with an initial speed
of 42 m/s directed at angle 0 = 600 above the horizontal. The
stone strikes A 5.5 s after launching. Find: h, the speed of the
stone just before impacting A, and the maximum height H reached
above the ground.
total time, t = 5.5 s
v0= 42 m/s
= 600
Let’s start by finding the v and y components of the initial velocity,
v0 (Note: this is always a good place to start!!)
opp v0 y
m
m

0
sin  

 v0 y  v0 sin    42  sin 60  36.4
hyp v0
s
s

adj v0 x
m
m

0
cos  

 v0 x  v0 cos    42  cos 60  21
hyp v0
s
s

v0 y
m
 36.4
s
v0 x
m
 21
s
Now we can proceed with
finding the height of the cliff!
total time, t = 5.5 s
v0= 42 m/s
v0y=36.4 m/s
= 600
v0x=21 m/s
v0 y  36.4
m
s
v0 x  21
m
s
1 2 
m
1
m
2
h  v0 y t  gt   36.4 5.5s   9.8 2 5.5s 
2
s
2
s 

 200.2m  148.2m  52m
Now let’s try and find the speed of the rock just before impact.
Remember that this will be the magnitude of the final velocity
vector. And this vector has both an x and y component.
total time, t = 5.5 s
v0= 42 m/s
vf
v0y=36.4 m/s
52m
= 600
v0x=21 m/s
Because there is no acceleration along the x axis: v0x=vfx=21 m/s
However in the y-direction there is acceleration so we must find
the y component of the final velocity:
m 
m
m

v fy  v0 y  gt   36.4    9.8 2 5.5s  17.5
s 
s 
s

vf 
v   v 
2
fy
2
fx
2
2
This negative
sign means the
y-component
is downward!
m  m
m

   17.5    21   27.3
s  s
s

total time, t = 5.5 s
v0= 42 m/s
vf=27.3 m/s
v0y=36.4 m/s
52m
= 600
v0x=21 m/s
Now let’s find the maximum height, H! To do this you have to know
that the instant the stone is at it’s maximum height the y component
of the velocity equals zero, vyMAV H = 0. Using this information we can
use:
v2fy  v02 y  2 gH
 0  v02 y  2 gH
2
2 gH  v02 y
m

36.4 

2
v0 y 
s
 H

 67.5m
m
2g
2  9.8 2
s
= 350
= 3.3 km
= 9.4 km
During volcano eruptions, chunks of solid rock can be blasted out of the
volcano; these projectiles are called volcanic bombs. At what initial speed
would a bomb have to be ejected, at angle 0=350 to the horizontal, from
the vent at A in order to fall at the foot of the volcano at B, at vertical
distance h=3.3 km and horizontal distance d=9.4 km ?
So we need to do is figure at what the initial speed of the
bombs need to be in order to hit point B.
v0=??
= 350
= 3.3 km
= 9.4 km
We have an equation for range, or horizontal distance:
x
x v 0 x t  v0 cos  0 t  v0 
cos  0 t
We have, x = 9.4 km, we have 0=350 what we need is the time, t, and
this is where it get tricky! Let’s start by using:
1 2
y  y0  h  v0 y t  gt
2
1 2
  h  v0 sin  0 t  gt
2
v0=??
= 350
= 3.3 km
= 9.4 km
1 2 I’m going to bring the –h over and make
 h  v0 sin  0 t  gt this a quadratic equation
2
1 2
 gt  v0 sin  0 t  h  0
2
a
b
c
a x b xc  0
2
What is the solution
to a quadratic
equation?
v0=??
= 350
= 3.3 km
= 9.4 km
b  4ac
a x  b x  c  0  x  b 
2a
a=½g
1 2
2
2

2
gt  v0 sin  0 t  h  0
b= v0 sin 0
c=h
1 
v0 sin  0   4 g h 
2 
t  v0 sin  0 
1 
2 g 
2 
2
this reduces
to
v0=??
= 350
= 3.3 km
= 9.4 km
1 
v0 sin  0   4 g h 
2 
t  v0 sin  0 
1 
2 g 
2 
now we can substitute this
2
v sin  0  2 gh
 v0 sin  0 
g
2
0
2
expression for time into
our range equation!
v0=??
= 350
= 3.3 km
= 9.4 km
x
v0 
cos  0 t
v02 sin 2  0  2gh
t  v0 sin  0 
g
Using a lot of algebra and tricks this equation becomes:
x
v0 
cos  0
g
m
 255.5
2x tan  0  h 
s
NOTE: I will ask you to show this on the exam!
v0=255 m/s
t= ?
= 350
= 3.3 km
= 9.4 km
Next, I would like to find the time of flight. We can rearrange this
equation to solve for time:
x
x
9400m
x  v0 x t  t 


 45 s
v0 x v0 cos  0 255 m cos 350
s
v0=255 m/s
t= 45 s
= 350
= 3.3 km
= 9.4 km
Finally how would air resistance change our initial velocity?
We expect the air to provide resistance but no appreciable lift to the
rock, so we would need a greater launching speed to reach the same target.