vectors and projectile motion chapter 3. adding vectors when adding vectors that fall on the same...
TRANSCRIPT
Vectors and Projectile Motion
Chapter 3
Adding Vectors
When adding vectors that fall on the same line, using pluses and minuses is sufficient.
When dealing with two dimensions, vector addition must be used. A complete answer includes magnitude, units, and a direction usually indicated by an angle.
Right Angle Vector AdditionC=resultant, means sum of two vectors.
2m
4m
=4.47m
c
In this case, Pythagorean theorem will solve for c.
In order to find , trig must be used. It is usually best to use the given sides, rather than the calculated side.
=63.4o
If you walk 2m north and then 4m east, what is your displacement?
Steps for Adding Perpendicular Vectors
Line vectors up head to tail.
Draw the resultant.
Find the resultant length using the Pythagorean theorem.
Use trig to find the angle.
Ensure that the angle measure has a distinct reference point, either by words or a picture.
Practice Problem
5m/s
10m/s
A sailboat experiences a 5m/s wind to the east while traveling 10m/s downstream. Find the resulting net velocity
5m/s
10m/s
Adding more than two vectors
Vectors can be added in any order.
Subtracting a vector is the same as adding a vector that is the same length, but going the exact opposite direction.
When drawing a picture, vectors may be rearranged into any order as long as their orientation is maintained.
Example
Practice
What is the total displacement of the motion, assuming each block is 1m long?
Vector directionDegrees on the unit circle
Start on the positive x and move CCW for positive degrees and CW for negative degrees.Examples, 40m @ 217°, 34m/s @ -73°
Compass direction27° S of WSouthwest would indicate a 45°Earth and space sciences use a system where north is 0° and one always has a positive angle and proceeds CW around the compass.
Adding vectors with random orientation
Components-A set of vectors that have a sum equal to a given vector.
Usually most useful to have components perpendicular to each other.
Breaking up a vector, called resolution, is the exact opposite of adding two perpendicular vectors to find a resultant.
Usually, components are along the x and y axis.
x
y
Component practice
50m@200°32m@67°
14m/s@-112° 6m/s@-300°
Steps for solving vector problems
Graphically represent the system to give yourself a check for your answer.Draw each vector individually.Using trig, find the x and y components of each vector.Add all the x components together and add all the y components together.Make a new diagram using the total x component and the total y component. Draw in the resultant and label a reference angle.Use the Pythagorean theorem and trig to find the resultant and angle
Example Problem
9m, 45o up from the horizontal
11m, 18o down from the horizontal
Splitting the Triangles Up
y1
9m, 45o up from the horizontal
x1
45o
11m, 18o down from the horizontal
x2
y2
18o
y1
x1
45o
x2
y2
18o
Working with the first triangle
945sin 1y
h
o
945cos 1x
h
a
my 36.645sin9 1
mx 36.645cos9 1 9m, 45o up from the horizontal
x1
45o
y1
Working with the second triangle
my 4.318sin11 2
1118cos 2x
h
a
1118sin 2y
h
o
mx 4.1018cos11 2 11m, 18o down from
the horizontal
x2
y2
18o
Putting everything together
myyy
mxxx
tot
tot
96.2
8.16
21
21
8.16
96.2tan
tot
tot
x
y
mc
mc
bac
1.17
9.291
8.1696.22
22222
xtot
ytot
c
01 1.108.16
96.2tan
Practice
Find the resultant of the following vectors:
450m @ 20o
360m @ 300o
290m @ 189o
405m @ 115o
Projectile Motion
Horizontal and Vertical Motion
Things to remember for 2D motion
Vertical and horizontal motion are independent.When something is traveling through the air, ignore the effects of air resistance.There is nothing pushing or pulling a projectile horizontally, therefore ax=0.For vertical motion, gravity is causing the vertical acceleration, so ay=-9.8m/s2.We will assume that projectiles landing at a height different from their initial height are always launched horizontally. Therefore, any initial velocity is an x piece. There is no y component for initial velocity.
Horizontal Components
tvx
tvtx
tvatx
ix
ix
ix
2
2
021
21
ixfx
ixfx
ixfx
vv
xvv
axvv
02
222
22
ixfx
ixfx
ixfx
vv
tvv
atvv
0
tvx
tvx
tvvx
tvvx
ix
ix
ixix
fxix
221
21
21
a=0
Vertical Components
2
2
2
21
021
21
gty
tgty
tvaty iy
gyv
gyv
ayvv
fy
fy
ify
2
20
22
22
gtv
gtv
atvv
fy
fy
ify
0
tvy
tvy
tvvy
f
fy
fyiy
21
021
21
Viy=0
a=g
Cliff Problems
A car drives off a 100m cliff at a speed of 47m/s. What is:
The time it takes to hit the ground?
It’s horizontal distance from the base of the cliff?
It’s final velocity?
Identify Vertical and Horizontal Components
Horizontal
vi=47m/s
vf=47m/s
a=0
t=?
x=?
Vertical
a=g=-9.8m/s2
vi=0
y=100m
t=?
vf=?
Part AUse y pieces because there is not enough information to solve for t using x components.
Since the car stops moving horizontally at the same time it stops moving vertically, the t found using the y components can be used for the x components as well.
tvaty iy 2
21
stg
y5.4
2
22t
a
y
Part B
tvatx i 2
21
mtvx i 3.212
Part C
The car is going both down and over at the end so vf has both x and y components.
vfx= vix
vfy must be calculated.
atvv if
sm
f atv 3.44
Part C Continued
sm
fyfx
c
cvv
6.64
222
01 3.43tan
tan
fx
fy
fx
fy
v
v
v
v
a
o
Put x and y components together and solve for both resultant and angle.
c
vfx
vfy
Practice Problems
A ball is thrown horizontally from the roof of a building 56m tall and lands 45m away from the base. What was the ball’s initial speed?
An onion runs off of a building at a speed of 22.2m/s and lands 36.0m from the base of the building. How tall is the building?
Starting and Ending at the Same Height
g
vR i 2sin2
g
vy i
2
sin 22
max
g
vt i sin2
Practice Problem
A water balloon is shot from a slingshot at an angle of 21o and with a velocity of 16m/s. What is:
The horizontal range?
The time it’s in the air?
It’s maximum height?
More Practice
A long jumper leaves the ground at 30o and travels 7.8m. How long is he in the air?
A pilot drops a flaming bag of poo from a plane in an attempt to hit the roof of the high school. If the plane is traveling at 160km/hr and is 160m above the ground, how far before he is directly overhead should the pilot drop the bag?