Motion in Two Dimensions

Holt Physics

Pages 98 – 106

Things to Remember

one-dimensional motion was either all in the horizontal (x) direction or all in the vertical (y) direction

acceleration of gravity (ag or g) is -9.8 m/s2 in the downward direction

gravity acts the SAMESAME on ALLALL objects!! downward velocity or position represented by “-“ find the components of any vector the same way as

previous section

Also remember…

Equations

Recall from chapter 2 when we studied 1-dimensional motion:

v = Δx/Δt (used with constant velocity)Δy = vΔt + ½ aΔt2

vf2 = vi

2 + 2aΔy

c2 = a2 + b2

Reviewing the Math #1

If you drop your cell phone from the top of your desk as you try to hide it (0.82 m), how long will it take to hit the ground?

Reviewing the Math #2

Find the two components of the velocity for each vector: (vx and vy)

v = 25 m/s horizontally

 

 

 

v = 25 m/s at 20° N of E

Chapter 3-3(Part 1)

Motion of Objects Projected Horizontally

Objective 1Understand how to separate a motion vector into its horizontal and vertical components

Objective 2Understanding that motions which are perpendicular to each other are also independent of each other

Objective 3Solve problems involving objects projected horizontally

Projectiles

Projectile:

objects thrown or launched into the air and subject to gravity

Projectile Motion:

motion through the air without a propulsion

Projectiles

Curved Motion:

path is parabolic (larger curve for higher launches)

Resolution into Components:

projectile motion can be separated into both vertical and horizontal components

Initial vertical component = viy = 0m/s

(object in free-fall, a = g = -9.8m/s2)

Initial horizontal component = vix = vx = shot at…

Components of a Horizontal Projectile

vx (horizontal velocity) is constant and there is no acceleration in this direction (we ignore air resistance)

vy is not constant, it is accelerated downward due to gravity (9.8 m/s2 down or -9.8 m/s2, where the - sign means down)

This means:

In the horizontal direction, equal distances are covered in equal amounts of time

In the vertical direction, there is accelerationacceleration, so different distances are covered in equal amounts of time

v0

x

y

x

y

x

y

x

y

x

y

x

y

•Object is in free-fall

•Acceleration is constant, and downward ay= g = -9.81m/s2

•The horizontal (x) component of velocity is constant

•The horizontal and vertical motions are independent of each other, but they have a common timeg = -9.81m/s2

WHAT DOES THIS MEAN?

Remember: vy = aΔt

What are the components at each second?

t = 0s 1s 2s 3s 4s

Horizontal and vertical velocities are completely independent of one another!

x

y

0

Frame of reference:

h

or

Δy

vi

Equations of motion:

horizontal (x)

uniform motion

vertical (y)

accelerated motion

acceleration ax = 0 ay = g = -9.81 m/s2

velocity vi = vx vy = ayΔt

vyf2 = 2ayΔy

displacement Δx = vxΔt Δy = ½ ayΔt2g or a

ASSUMPTIONS:

• x-direction (horizontal): uniform motion

• y-direction (vertical): accelerated motion

• no air resistance

POSSIBLE QUESTIONS:

• What is the total time of the motion?

• What is the horizontal range (Δx)?

• What is the initial velocity?

• What is the final velocity?

Analysis of Motion

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

How long does it take the football to reach the bottom of the cliff?

78.4 m

vx= 5m/s

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

How long does it take the football to reach the bottom of the cliff?

78.4 m

vx= 5m/s

Unknown:Unknown: Δt

Given:Given:yf = 78.4myi = 0 mvyi = 0 m/svxi = 5.0 m/sa = -9.8m/s2

Equation:Equation:Δy = ½ aΔt2

SolvingSolving for ΔtΔt2 = 2 Δyf/aΔt = √ 2Δyf/a Δt = √ 2(-78.4m)/(-9.8m/s2)Δt = 4.0 s

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

How far from the base of the cliff does the football land?

78.4m

vx= 5m/s

?

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

How far from the base of the cliff does the football land?

78.4m

vx= 5m/s

Unknown:Unknown: Δx

Given:Given:yf = 78.4myi = 0 mvyi = 0 m/svxi = 5.0 m/sa = -9.8 m/s2

Δt = 4.0 s

Equation:Equation:Δ x = vx Δt

SolvingSolving for Δ xΔ x = (5.0m/s)(4.00s)Δ x = 20. m

?

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)?

78.4m

vx= 5m/s

vf

A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high

What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)?

78.4m

vx= 5m/s

Unknown:Unknown: vfx , vfy

Given:Given:vxi =5.0 m/s yi = 0 m

yf = 78.4mvyi = 0 m/s

ΔΔt = 4.0 st = 4.0 sa = -9.8 m/sa = -9.8 m/s22

Equation Equation for vxf

vxf = vxi

Solving Solving vxf = 5.0 m/s

EquationEquation for vyf

vyf = a Δt

Solving Solving vyf = (-9.8 m/s2 )(4s) vyf = -39 m/s

vf

Sample Problem

Someone is being chased down a river by someone else in a faster craft. Just as the fast boat pulls up to the slower boat, both reach the edge of a 5.0 m waterfall. If the slower boat’s velocity is 15 m/s and the faster boat’s speed is 26 m/s, how far apart will the two vessels be when they land?

sketch known(x) known(y) unknown

known(x) known(y) unknown

equation/solution

equation/solution

Sample Problem

An African Spitting Cobra can raise its head straight up approximately 0.61m. An average distance that the poisonous spit travels is 3.60m. What is the horizontal velocity of this deadly venom?

sketch known(x) known(y) unknown

equation/solution

Sample Problem

A girl jumps off of the 10. m platform with a horizontal velocity of 2.0 m/s. How far from the end of the platform does she hit the water?

(Δt = 1.43 sec)

Sample Problem

An army helicopter needs to drop supplies to troops in the field. If the army helicopter is flying at an altitude of 500 m and a horizontal velocity of 10 m/s, how far before it gets to the target zone should they drop the supplies?

(Δx = 101m)