Position and Displacement we want to locate a point in space we can use a position vect t extends from a reference point to the location we are tryin describe. k z j y i x r 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) k j i r 5 2 3 the displacement is just the change in a particle’s position
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. i, j, k are the unit vectors along the x, y, z, directions x, y, z are the components of the vector along those - PowerPoint PPT Presentation
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Position and Displacement
If we want to locate a point in space we can use a position vectorthat extends from a reference point to the location we are tryingto describe.
kzjyixr
i, j, k are the unit vectorsalong the x, y, z, directions
x, y, z are the componentsof the vector along thosedirections (i, j, k)
kjir
523 the displacement is just the change in a particle’s position
kzjyixr iiii
If we have an initial and final positions described by two positionvectors, ri and rf
kzjyixr ffff
So our displacement is defined as:
kzzjyyixx
rrr
ififif
if
Let’s look at an example:
kmjmimr
kmjmimr
f
i
829
523
kmjmimr
kmjmimr
f
i
829
523
kmim
kmmjmmimm
rrr if
312
582239
the definitions for velocity and acceleration are basicallythe same but you have to remember that these are vectorsquantities:
average velocity:t
rr
t
rv ifAvg
referring to the previous example, if it took the particle3 seconds to to the displacement, then the average velocityis:
ks
mis
mki
t
rvAvg
16
3312
similarly, the instantaneous velocity is:
td
rdv
GO TO HITT
acceleration follows the same rrr-guments:
t
vv
t
va ifAvg
instantaneous acceleration is:
td
vda
the acceleration vector shows thedirection of the acceleration of theparticle
Motion in more than one dimension.http://www.physicsclassroom.com/mmedia/vectors/mzi.html
straight line(line of site)
path ofbullet
path of monkey
the monkey begins to fall ad theprecise moment when the ballleaves the barrel of the gun
The ball and themonkey arrive atthe point markedby the reddot at thesame time
Monkey: Object under constant acceleration in 1 – dimension
Bullet: The motion of the bullet is a combinationof motion with a constant velocity (alongthe x-axis) and motion with constant acceleration (along the y-axis)
y
x
x
y
jvivv iyixi
vi
viy opp
vix adj
coscos
sinsin
iixi
ix
iiyi
iy
vvv
v
hyp
adj
vvv
v
hyp
opp
|v i| h
yp
To start analyzing this problem we must find the x, and y componentsof the velocity:
Let’ start by looking at an example where the launch angle is zero
h = 1 mv0X = 2 m/s = 0
x = ?
g
ht
g
httgh
ttimeforsolvetgh
tgtgtvhv
tgtvh
tvx
yy
y
x
222
21
21
21
0
21
22
2
2200
20
0
,
now that we have an expression for time, t we can find the range x
h = 1 mv0X = 2 m/s = 0
x = ?
m
smm
s
m
g
hvx
g
ht
tvx
x
x
9089
122
2
2
2
0
0
..
We could also find the final velocity, vf
vf
??
fy
xfx
fyfxf
fyfxf
vs
mvv
vvv
yvxvv
20
22
s
m
ss
m
gtvv yfy
54
450890 2
0
.
..
s
m
s
m
s
m
vvv fyfxf
84
54222
22
.
.
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 harborentrance 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?
tv
tvx x
cos0
0
We need an expression for the time, t
tv
tvx x
cos0
0
We need an expression for the time, t
200 2
1gttvyyh y since the projectile will
return to the same height that it left from,h = y –y0 = 0 so our equation becomes:
20
20 2
121
0 gttvgttv y sin
20 2
10 gttv sin from here we can solve for time, t
factor out a t, and it cancels: sinsin 00 21
21
0 vgtgtv
g
vtvgt
sinsin 0
0
22 now substitute this back into:
tvx cos0
sincossin
coscosg
v
g
vvtvx
200
00
22
from trigonometry: cossinsin 22
220 sing
vx this is called the Range equation!
DANGER WILL ROBINSON: This equation only works if the projectilereturns to the same height that it was launched!!!
8160
82
5608922 2
2
20
20
.
/
/.sinsin
sm
msm
v
gx
g
vx
oo or
or
6327
3125654180654
81602 1
...
.sin
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 speedof 42 m/s directed at angle 0 = 600 above the horizontal. Thestone strikes A 5.5 s after launching. Find: h, the speed of thestone just before impacting A, and the maximum height H reachedabove the ground.
v0= 42 m/s
= 600
total time, t = 5.5 s
Let’s start by finding the v and y components of the initial velocity,v0 (Note: this is always a good place to start!!)
s
m
s
mvv
v
v
hyp
adj
s
m
s
mvv
v
v
hyp
opp
xx
yy
216042
4366042
000
0
0
000
0
0
coscoscos
.sinsinsin
s
mv
s
mv xy 21436 00 . Now we can proceed with
finding the height of the cliff!
mmm
ss
ms
s
mgttvh y
5221482200
558921
5543621 2
22
0
..
....
v0= 42 m/s
= 600
total time, t = 5.5 s
s
mv
s
mv xy 21436 00 .
v0y=36.4 m/s
v0x=21 m/s
Now let’s try and find the speed of the rock just before impact.Remember that this will be the magnitude of the final velocityvector. And this vector has both an x and y component.
v0= 42 m/s
= 600
total time, t = 5.5 s
v0y=36.4 m/s
v0x=21 m/s
52m
Because there is no acceleration along the x axis: v0x=vfx=21 m/sHowever in the y-direction there is acceleration so we must findthe y component of the final velocity:
s
ms
s
m
s
mgtvv yfy 5175589436 20 ....
This negativesign means they-componentis downward!
vf
s
m
s
m
s
mvvv fxfyf 32721517
2222 ..
v0= 42 m/s
= 600
total time, t = 5.5 s
v0y=36.4 m/s
v0x=21 m/s
52m
vf=27.3 m/s
Now let’s find the maximum height, H! To do this you have to knowthat the instant the stone is at it’s maximum height the y componentof the velocity equals zero, vyMAV H = 0. Using this information we canuse:
m
smsm
g
vHvgH
gHvgHvv
yy
yyfy
567892
436
22
202
2
2
202
0
20
20
2
..
.
During volcano eruptions, chunks of solid rock can be blasted out of thevolcano; these projectiles are called volcanic bombs. At what initial speedwould 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 verticaldistance h=3.3 km and horizontal distance d=9.4 km ?
= 350
So we need to do is figure at what the initial speed of thebombs need to be in order to hit point B.
= 9.4 km
= 3.3 km
= 350
= 9.4 km
= 3.3 km
v0=??
We have an equation for range, or horizontal distance:
t
xvtvtvx x
00000
cos
cos
We have, x = 9.4 km, we have 0=350 what we need is the time, t, andthis is where it get tricky! Let’s start by using:
200
200 2
121
gttvhgttvhyy y sin
= 350
= 9.4 km
= 3.3 km
v0=??
200 2
1gttvh sin I’m going to bring the –h over and make
this a quadratic equation
021
002 htvgt sin
a b c
02 cxbxaWhat is the solutionto a quadraticequation?
= 350
= 9.4 km
= 3.3 km
v0=??
a
acbbxcxbxa
24
02
2
021
002 htvgt sin
a = ½ gb= v0 sin 0
c = h
g
hgv
vt
21
2
21
4200
00
sin
sin
this reducesto
= 350
= 9.4 km
= 3.3 km
v0=??
g
ghvv
g
hgv
vt
2
21
2
21
4
022
000
200
00
sinsin
sin
sin
now we can substitute this expression for time into our range equation!
= 350
= 9.4 km
= 3.3 km
v0=??
t
xv
00 cos
g
ghvvt
2022
000
sinsin
Using a lot of algebra and tricks this equation becomes:
s
m
hx
gxv 5255
2 000 .
tancos
NOTE: I will ask you to show this on the exam!
= 350
= 9.4 km
= 3.3 km
v0=255 m/s
Next, I would like to find the time of flight. We can rearrange thisequation to solve for time:
s
sm
m
v
x
v
xttvx
xx 45
35255
94000000
0 coscos
t= ?
= 350
= 9.4 km
= 3.3 km
v0=255 m/s
t= 45 s
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.