Introduction to 2-Dimensional Motion

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# Introduction to 2-Dimensional Motion - PowerPoint PPT Presentation

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.

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## Introduction to 2-Dimensional Motion

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### 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.
• How far west have you traveled in 2.5 minutes?
• 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.
• How far west have you traveled in 2.5 minutes?
• 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.
• How far west have you traveled in 2.5 minutes?
• How far south have you traveled in 2.5 minutes?

v = 5 m/s, Q = 40o,

t = 2.5 min = 150 s

vx= v cosQvy= v sin Q

vx= 5 cos 40vy= 5 sin 40

vx= vy=

x = vxt y = vyt

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

x = y =

vx

vy

v = 5 m/s

Sample Problem

A 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.

• What are the x and y positions at 5.0 seconds?
Sample Problem

A 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.

• 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 + at

x = 0 + (-4.4)(5) y = 6.2 + (0)(5)

x = y =

Sample Problem

A 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.

• What are the x and y components of velocity at this time?
Sample Problem

A 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.

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

vx = vo,x+ axtvy = 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
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,y2 - 2g(y – yo)

### 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.

vo

Zero Launch angle
• A zero launch angle implies a perfectly horizontal launch.
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 = ?

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,y2 - 2g(y – yo)

Vy2 = (0)2 – 2(9.8)(0 – 108)

Vy =

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,y2 - 2g(y – yo)

v = Vy2 = (0)2 – 2(9.8)(0 – 108)

Vy =

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,xt

t = x/vo,x

t = 8.95/6.75

t =

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.
• How far were you from the second baseman?
• What is the distance of the vertical drop?

Should be able to do this on your own!

### General Launch Angle Projectiles

vo

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

vo

General launch angle
• You must begin problems like this by resolving the velocity vector into its components.

Vo,y = Vo sin 

Vo,x = Vo cos 

Resolving the velocity
• Use speed and the launch angle to find horizontal and vertical velocity components

Vo

Vo,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.

Vo

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, q = 25o, g = -9.8 m/s2, Remember: because it lands at the same height: Dy = 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.

y

x

Trajectory of a 2-D Projectile
• Definition: The trajectory is the path traveled by any projectile. It is plotted on an x-y graph.

y

x

Trajectory of a 2-D Projectile
• Mathematically, the path is defined by a parabola.

y

x

Trajectory of a 2-D Projectile
• For a projectile launched over level ground, the symmetry is apparent.

y

x

Range of a 2-D Projectile
• Definition: The RANGE of the projectile is how far it travels horizontally.

Range

y

x

Maximum height of a projectile
• The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward.

Maximum

Height

Range

y

x

Maximum height of a projectile
• The vertical velocity component is zero at maximum height.

Maximum

Height

Range

y

x

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

Maximum

Height

Range

y

g

g

g

g

g

x

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

y

x

Velocity of a projectile
• Velocity is tangent to the path for the entire trajectory.

v

v

v

vo

vf

y

x

Velocity of a projectile
• The velocity can be resolved into components all along its path.

vx

vx

vy

vy

vx

vy

vx

vy

vx

y

x

Velocity of a projectile
• Notice how the vertical velocity changes while the horizontal velocity remains constant.

vx

vx

vy

vy

vx

vy

vx

vy

vx

y

x

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

vx

vx

vy

vy

vx

vy

vx

vy

vx

vo

-

vo

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

t

to = 0

Time of flight for a projectile
• The projectile spends half its time traveling upward…

t

to = 0

2t

Time of flight for a projectile
• … and the other half traveling down.

### Projectile Lab

Projectile Lab

The 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:
• a table of data
• a graph of the trajectory
• a calculation of the launch velocity of the ball obtained from the data
• Hints and tips:
• The thin paper strip is pressure sensitive. Striking the paper produces a mark.
• 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.
• 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 = vo2sin(2q)/g.
• R: range of projectile fired over level ground
• vo: initial velocity
• g: acceleration due to gravity
• q: launch angle
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.

How far from the golfer does the ball land?

Vo = 42m/s, q = 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.
• 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?