Introduction to 2-Dimensional Motion. 2-Dimensional Motion Definition: motion that occurs with both x and y components. Example: Playing pool. Throwing

Introduction to 2-Dimensional Motion

2-Dimensional Motion Definition: motion that occurs with

both x and y components. Example:

Playing pool . Throwing a ball to another person.

Each dimension of the motion can obey different equations of motion.

Solving 2-D Problems Resolve all vectors into components

x-component Y-component

Work the problem as two one-dimensional problems. Each dimension can obey different

equations of motion. Re-combine the results for the two

components at the end of the problem.

Sample Problem You run in a straight line at a speed of 5.0 m/s in a

direction that is 40o south of west.a) How far west have you traveled in 2.5 minutes?b) How far south have you traveled in 2.5 minutes?

Sample Problem You run in a straight line at a speed of 5.0 m/s in a

direction that is 40o south of west.a) How far west have you traveled in 2.5 minutes?b) How far south have you traveled in 2.5 minutes?

v = 40 m/s

Sample Problem You run in a straight line at a speed of 5.0 m/s in a direction that is 40o south of west.

a) How far west have you traveled in 2.5 minutes?b) How far south have you traveled in 2.5 minutes?

v = 5 m/s, = 40o, t = 2.5 min = 150 svx = v cosvy= v sin

vx = 5 cos 40vy= 5 sin 40

vx = vy=

x = vx t y = vyt

x = ( )(150) y = ( )(150)x = y =

v = 5 m/s

vx

vy

Sample ProblemA particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2

.

a) What are the x and y positions at 5.0 seconds?


.

a) What are the x and y positions at 5.0 seconds?

Vo,y = 6.2 m/s, Vo,x = 0 m/s, t = 5 s, ax = -4.4 m/s2, ay = 0 m/s2

x = ? y = ?x = vo,x + at y = vo,y + atx = 0 + (-4.4)(5) y = 6.2 + (0)(5)x = y =


.

b) What are the x and y components of velocity at this time?


.

b) What are the x and y components of velocity at this time?

vx = vo,x + axt vy = voy + axt

vx = 0 + (-4.4)(5)vy = 6.2 + 0(5)

vx = vy =

Projectiles

Projectile Motion Something is fired, thrown, shot, or

hurled near the earth’s surface. Horizontal velocity is constant. Vertical velocity is accelerated. Air resistance is ignored.

1-Dimensional Projectile Definition: A projectile that moves in a

vertical direction only, subject to acceleration by gravity.

Examples: Drop something off a cliff. Throw something straight up and catch it.

You calculate vertical motion only. The motion has no horizontal

component.

2-Dimensional Projectile Definition: A projectile that moves both

horizontally and vertically, subject to acceleration by gravity in vertical direction.

Examples: Throw a softball to someone else. Fire a cannon horizontally off a cliff. Shoot a monkey with a blowgun.

You calculate vertical and horizontal motion.

Horizontal Component of Velocity Is constant Not accelerated Not influence by gravity Follows equation: x = Vo,xt

Horizontal Component of Velocity

Vertical Component of Velocity Undergoes accelerated motion Accelerated by gravity (9.8 m/s2

down) Vy = Vo,y - gt y = yo + Vo,yt - 1/2gt2

Vy2 = Vo,y

2 - 2g(y – yo)

Horizontal and Vertical

Horizontal and Vertical

Zero Launch Angle Projectiles

Launch angle Definition: The angle at which a

projectile is launched. The launch angle determines what

the trajectory of the projectile will be. Launch angles can range from -90o

(throwing something straight down) to +90o (throwing something straight up) and everything in between.

Zero Launch angle

A zero launch angle implies a perfectly horizontal launch.

vo

Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are

approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is

flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

yo = 108 m, y = 0 m, g = -9.8 m/s2, vo,x = 3.6 m/s

v = ?

2y

2x vvv

Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m

high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s

v = ? Gravity doesn’t change horizontal velocity. vo,x = vx =

3.6 m/s

Vy2 = Vo,y

2 - 2g(y – yo)

Vy2 = (0)2 – 2(9.8)(0 – 108)

Vy =

2y

2x vvv

Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m

high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s

v = ? Gravity doesn’t change horizontal velocity. vo,x = vx =

3.6 m/s

Vy2 = Vo,y

2 - 2g(y – yo)

v = Vy2 = (0)2 – 2(9.8)(0 – 108)

Vy =

2y

2x vvv

22 ) ((3.6)v

Sample Problem An astronaut on the planet Zircon tosses a rock horizontally

with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?

Sample Problem An astronaut on the planet Zircon tosses a rock horizontally with a

speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?

vo,x = 6.75 m/s, x = 8.95 m, y = 0 m, yo = 1.2, Vo,y

= 0 m/s

g = ?y = yo + Vo,yt - 1/2gt2

g = -2(y - yo - Vo,yt)/t2

g = -2[0 – 1.2 – (0)( )]/( )2

g =

x = vo,xtt = x/vo,x t = 8.95/6.75t =

Sample Problem Playing shortstop, you throw a ball horizontally to the second

baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later.

a) How far were you from the second baseman?b) What is the distance of the vertical drop?

Should be able to do this on your own!

General Launch Angle Projectiles

General launch angle

vo

Projectile motion is more complicated when the launch angle is not straight up or down (90o or –90o), or perfectly horizontal (0o).

General launch angle

vo

You must begin problems like this by resolving the velocity vector into its components.

Resolving the velocity Use speed and the launch angle to find

horizontal and vertical velocity components

VoVo,y = Vo sin

Vo,x = Vo cos

Resolving the velocity Then proceed to work problems just like

you did with the zero launch angle problems.

VoVo,y = Vo sin

Vo,x = Vo cos

Sample problem A soccer ball is kicked with a speed of 9.50 m/s at an

angle of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?

Sample problem A soccer ball is kicked with a speed of 9.50 m/s at an angle

of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?

vo = 9.5 m/s, = 25o, g = -9.8 m/s2, Remember: because it lands at the same height: y = y – yo = 0 m and vy =- vo,y

Find: Vo,y = Vo sin and Vo,x = Vo cos

Vo,y = 9.5 sin 25 Vo,x = Vo cos

Vo,y = Vo,x =

t = ?Vy = Vo,y - gt

t = (Vy - Vo,y )/g

t = [( ) – ( )]/9.8 don’t forget vy =- vo,y

t =

Sample problem Snowballs are thrown with a speed of 13 m/s from a

roof 7.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25o above the horizontal. When the snowballs land, is the speed of A greater than, less than, or the same speed of B? Verify your answer by calculation of the landing speed of both snowballs.

We’ll do this in class.

Projectiles launched over level ground These projectiles have highly

symmetric characteristics of motion. It is handy to know these

characteristics, since a knowledge of the symmetry can help in working problems and predicting the motion.

Lets take a look at projectiles launched over level ground.

Trajectory of a 2-D Projectile

x

y

Definition: The trajectory is the path traveled by any projectile. It is plotted on an x-y graph.


x

y

Mathematically, the path is defined by a parabola.


x

y

For a projectile launched over level ground, the symmetry is apparent.

Range of a 2-D Projectile

x

y

Range

Definition: The RANGE of the projectile is how far it travels horizontally.

Maximum height of a projectile

x

y

Range

MaximumHeight

The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward.


x

y

Range

MaximumHeight

The vertical velocity component is zero at maximum height.


x

y

Range

MaximumHeight

For a projectile launched over level ground, the maximum height occurs halfway through the flight of the projectile.

Acceleration of a projectile

g

g

g

g

g

x

y

Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.

Velocity of a projectile

vo

vf

v

v

v

x

y

Velocity is tangent to the path for the entire trajectory.


vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

The velocity can be resolved into components all along its path.


vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

Notice how the vertical velocity changes while the horizontal velocity remains constant.


vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

Maximum speed is attained at the beginning, and again at the end, of the trajectory if the projectile is launched over level ground.

vo -

vo


Launch angle is symmetric with landing angle for a projectile launched over level ground.

to = 0

t

Time of flight for a projectile

The projectile spends half its time traveling upward…

Time of flight for a projectile

to = 0

t

2t

… and the other half traveling down.

Position graphs for 2-D projectiles

x

y

t

y

t

x

Velocity graphs for 2-D projectiles

t

Vy

t

Vx

Acceleration graphs for 2-D projectiles

t

ay

t

ax

Projectile Lab

Projectile LabThe purpose is to collect data to plot a trajectory for a projectile launched horizontally, and to calculate the launch velocity of the projectile. Equipment is provided, you figure out how to use it. What you turn in:

1. a table of data 2. a graph of the trajectory3. a calculation of the launch velocity of the

ball obtained from the data Hints and tips:

1. The thin paper strip is pressure sensitive. Striking the paper produces a mark.

2. You might like to hang a sheet of your own graph paper on the brown board.

More on Projectile Motion

The Range Equation Derivation is an important part of

physics. Your book has many more

equations than your formula sheet. The Range Equation is in your

textbook, but not on your formula sheet. You can use it if you can memorize it or derive it!

The Range Equation R = vo

2sin(2)/g. R: range of projectile fired over level

ground vo: initial velocity g: acceleration due to gravity : launch angle

Deriving the Range Equation

Sample problemA golfer tees off on level ground, giving the ball an initial speed of

42.0 m/s and an initial direction of 35o above the horizontal.

How far from the golfer does the ball land?

Vo = 42m/s, = 35o, g = 9.8 m/s2

R = ?

Sample problem A golfer tees off on level ground, giving the ball an

initial speed of 42.0 m/s and an initial direction of 35o above the horizontal.

b) The next golfer hits a ball with the same initial speed, but at a greater angle than 45o. The ball travels the same horizontal distance. What was the initial direction of motion?

Documents

Introduction to 2-Dimensional Motion. 2-Dimensional Motion Definition: motion that occurs with both x and y components. Example: Playing pool. Throwing