2.4 Motion in Two Dimensions

2.4 Motion in Two Dimensions2.4 Motion in Two Dimensions

StandardStandard

SP1. Students will analyze the relationships between force,

mass, gravity, and the motion of objects.

f. Measure and calculate two-dimensional motion (projectile

and circular) by using component vectors.

Projectile -Any object which projected by some means and continues to move due to its own inertia (mass). Motion is affected by gravity. It is a projectile from the moment it is launched until the instant before it hits the ground

2.4 Motion in Two Dimensions2.4 Motion in Two DimensionsMeasure and calculate two-dimensional motion by using component vectors.

Projectiles Move in Two Projectiles Move in Two DimensionsDimensions

Since a projectile moves in 2-dimensions, Since a projectile moves in 2-dimensions, it therefore has 2 components just like a it therefore has 2 components just like a resultant vector.resultant vector.

Horizontal and VerticalHorizontal and Vertical

Acceleration is caused by gravity

In what axis?

Only in the y

What happens in the x?

Constant velocity

t

dv

x

x


So if there is no gravity


The actual pathway has constant x velocity and changing y

Acceleration in the -y


Horizontal Velocity ComponentHorizontal Velocity Component

NEVER changes, NEVER changes, covers equal covers equal displacements in displacements in equal time periods.equal time periods.

This means the initial This means the initial horizontal velocity horizontal velocity equals the final equals the final horizontal velocityhorizontal velocity

In other words, the In other words, the horizontal velocity horizontal velocity is CONSTANT. BUT is CONSTANT. BUT WHY?WHY?

Gravity DOES NOT Gravity DOES NOT work horizontally to work horizontally to increase or increase or decrease the decrease the velocity.velocity.

In the y axis, acceleration is always -9.80m/s2

All the acceleration equations apply

In the y axis motion is identical to falling

dy = vi y t+ ½ at2

v f y= vi y + at

vf y2= vi y

2 + 2ad


Vertical Velocity ComponentVertical Velocity Component Changes (due to gravity), does Changes (due to gravity), does NOT NOT

cover cover equal displacements in equal time equal displacements in equal time periods.periods.

Combining the ComponentsCombining the Components

Together, these Together, these components produce components produce what is called a what is called a trajectory or path. trajectory or path.

This This path is path is parabolic in parabolic in nature.nature.

Projectile Motion: Basic EquationsProjectile Motion: Basic EquationsMeasure and calculate two-dimensional motion by using component vectors.

We will assume

no air resistance

gravity is

-9.80 m/s2

the Earth is not moving (Frame of Reference)

XX YY

vvxx = = vvoyoy = =

ddxx = = vvyy = =

t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = =

That leaves the following variables

2.5 Projectile Motion: Basic Equations2.5 Projectile Motion: Basic EquationsMeasure and calculate two-dimensional motion by using component vectors.

The two axes are independent of each other except for time

XX YY

vvxx = = vvoyoy = =

ddxx = = vvyy = =

t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = =

2.5 Projectile Motion: Basic Equations2.5 Projectile Motion: Basic EquationsMeasure and calculate two-dimensional motion by using component vectors.

2.6 Zero Launch Angle2.6 Zero Launch AngleMeasure and calculate two-dimensional motion by using component vectors.

Projectiles which have NO upward trajectory and NO initial VERTICAL velocity.


If an object is launched at 0o

So our chart becomes

XX YY

vvxx = = vv v v i i yy = = 00

ddxx = = v v f f yy = =

t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = =


Now we can fill in whatever else is given in the problem

XX YY

vvxx = = vv v v ii yy = = 00


t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = =


Horizontally Launched Horizontally Launched ProjectilesProjectiles

To analyze a projectile in 2 dimensions we need To analyze a projectile in 2 dimensions we need 2 equations. One for the x direction and one for 2 equations. One for the x direction and one for the y direction. And for this we use 2 kinematic the y direction. And for this we use 2 kinematic equationequation

Horizontal = vHorizontal = v x x = d = dxx/t Vertical = d/t Vertical = dyy= ½ at= ½ at22

Horizontal = v x = d/t Vertical = d= ½ at2

Remember the velocity is CONSTANT horizontally, so that means the acceleration is ZERO! Use constant velocity formula.

Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.

A person on a skate board with a constant speed of 1.3 m/s releases a ball from a height of 1.25 m above the ground.

What variable can you

fill in?XX YY

vvxx = = vv v v i i yy = = 00


t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = =


A person on a skate board with a constant speed of 1.3 m/s releases a ball from a height of 1.25 m above the ground.

What variable can you

fill in?XX YY

vvxx = = 1.3m/s1.3m/s v v i i yy = = 00


t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 m


We are now prepared to answer some questions

A) How long is the ball in the air?

XX YY

vvxx = = 1.3m/s1.3m/s v v i i yy = = 00


t =t = t =t =

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 mst

a

dt

atd

attvd

y

y

y

505.080.9

)25.1(2

2

221

221

0


We are no prepared to answer some questions

A) How long is the ball in the air?

XX YY

vvxx = = 1.3m/s1.3m/s v v i i yy = = 00


t = t = 0.505s0.505s t = t = 0.505s0.505s

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 m


st

a

dt

atd

attvd

y

y

y

505.080.9

)25.1(2

2

221

221

0

Basically a one dimensional problem

Notice that time is the same in both the Y and the X

XX YY

vvxx = = 1.3m/s1.3m/s vvi i yy = = 00

ddxx = = vvf f yy = =

t = t = 0.505s0.505s t = t = 0.505s0.505s

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 m


st

a

dt

atd

attvd

y

y

y

505.080.9

)25.1(2

2

221

221

0

We can now solve for X variable if we want

B) What is the x displacement?

XX YY

vvxx = = 1.3m/s1.3m/s v v ii yy = = 00

ddxx = = vvff yy = =

t = t = 0.505s0.505s t = t = 0.505s0.505s

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 mmd

d

tvdt

dv

x

x

xx

x

x

656.0

)505.0)(3.1(


We can now solve for distance in the x direction if we want

B) What is the x displacement?

XX YY

vvxx = = 1.3m/s1.3m/s v v i i yy = = 00

ddxx = =0.656m0.656m v v f f yy = =

t = t = 0.505s0.505s t = t = 0.505s0.505s

a = -9.80 m/sa = -9.80 m/s22

ddyy = = -1.25 m-1.25 mmd

d

tvdt

dv

x

x

xx

x

x

656.0

)505.0)(3.1(


ExampleExample A plane traveling with a horizontal velocity of 100 m/s is A plane traveling with a horizontal velocity of 100 m/s is

500 m above the ground. At some point the pilot decides to 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. drop some supplies to designated target below.

(a) How long is the drop in the air?(a) How long is the drop in the air? (b) How far away from point where it was launched will it (b) How far away from point where it was launched will it

land?land?

ExampleExample A plane traveling with a horizontal A plane traveling with a horizontal

velocity of 100 m/s is 500 m above velocity of 100 m/s is 500 m above the ground. At some point the pilot the ground. At some point the pilot decides to drop some supplies to decides to drop some supplies to designated target below. designated target below.

(a) How long is the drop in the air?(a) How long is the drop in the air? (b) How far away from point where it (b) How far away from point where it

was launched will it land?was launched will it land?

ddyy = ½ at = ½ at22

-500 = ½ (-9.8)t-500 = ½ (-9.8)t22

t = 10.1 st = 10.1 s Then:Then:

vvxx = d = dxx/t/t

100 = d100 = dxx/10.1/10.1

ddxx = 1010 m = 1010 m

What do I know?

What I want to know?

v x = 100 m/s t = ?

dy = - 500 m dx = ?

vi y = 0 m/s

a = - 9.80 m/s2

2.7 General Launch Angle2.7 General Launch AngleMeasure and calculate two-dimensional motion by using component vectors.

Vertical Launch AngleVertical Launch Angle

2.4 Motion in Two Dimensions

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