scrambled eggs

see eye to eyeup to no good

pipe down or downpipe

3.2 Equations of Kinematics in Two Dimensions

• Equations of Kinematics

tvvx o 21

221 attvx o

atvv o

axvv o 222

3.2 Equations of Kinematics in Two Dimensions – the horizontal component

tavv xoxx tvvx xox 21

xavv xoxx 222 221 tatvx xox

3.2 Equations of Kinematics in Two Dimensions – the vertical component

tavv yoyy

221 tatvy yoy

tvvy yoy 21

yavv yoyy 222

3.2 Equations of Kinematics in Two Dimensions• The x part of the motion occurs exactly as it

would if the y part did not occur at all, and vice versa.

Projectile MotionChapter 3Section 3

What is a projectile?

• When a long jumper approaches his jump, he runs along a straight line which can be called the x-axis.

• When he jumps, as shown in the diagram below, his velocity has both horizontal and vertical components

What is a projectile?• Projectiles are objects in free fall that have an

initial horizontal velocity.• Free fall implies that the only force acting on the object

is gravity.• Initial horizontal velocity results in motion in two

dimensions.

• Examples of projectiles include• An arrow flying towards a target• A baseball thrown in the air at an angle other than 90o,

unless they are thrown straight up.

Projectile motion

• Projectiles follow parabolic trajectories.• Trajectory is the path of a flying object.

• The horizontal and vertical motions are independent.• We can write separate equations of motion for each

direction. • Horizontal motion on the x-axis and • the vertical motion on the y-axis

What is a projectile?• If a ball is thrown horizontally, (yellow) and

another ball is dropped (red)-----they will both• HIT THE GROUND AT THE SAME TIME!!!!!!!!• We shall do a demo of this

Solving Projectile Motion Problems

• You need to separate the motion down into 2 parts!

• The first part is the horizontal movement• The second part is the vertical movement• Keep the two sets of information separate

using a table when solving such problems

Horizontal motion equations and constants

ΔtvΔx ox

projectile a ofcomponent vertical theaffectsonly gravity 0

)( 221

x

xox

a

tatvx

xoxx

xoxx

vvv

tavv

constant xv

Vertical motion equations and constants

Δt

vvg oyy

tgvv oyy

221

sm

221

))((

81.9

)(

2

tgtvy

ga

tatvy

oy

y

yoy

Calculating time of fall• If a projectile is launched horizontally, this tells

us that the object has an initial horizontal velocity (ie: in the x-axis)

• For such projectiles, the initial vertical velocity is zero (ie. in the y-axis) (voy=0)

g

yt

tgy

tgtvy oy

2

))((

))((2

21

221

Example 1

• People in movies often jump from buildings into pools. If a person jumps from the 10th floor (30.0 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump?

Step 1: Write down Givens and unknowns.

• You will have to separate your y-components from your x-components using a table

X- Component Y - Component

Dx = 5.0 m (from building to pool) Dy = -30.0 m (negative due to downward direction of fall)

vox = ? voy = 0 (horizontal projectile)

ax = 0 ay = -9.81 m/s2

Step 2: Write down equations that might solve it.

g

yt

2

• We can use the y equation to solve for the time of the jump.

• We then use this calculated time in the x equation to solve for the initial velocity.

• These are the two common equations you will use in this section!!!

tvx ox

Step 3: Carry out the solution.Time for the jump

• Using the y equation to solve for the time of the jump.

g

yt

2

s 47.2 9.81-

) 2(-30

t

t

Step 4: Carry out the solution.Using our time to calculate the initial horizontal velocity

• Using the x equation to solve for the initial horizontal velocity.

m/s 0.2s 47.2

m 0.5

ox

ox

vt

x

tvx

Practice Problem 1

• An autographed baseball rolls off of a 0.70m high desk. And strikes the floor 0.25m away from the desk. How fast did it roll off the table?

Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)

X- Component Y - Component

Dx = 0.25 m (from base of desk) Dy = -0.7 m (negative due to downward direction of fall)

vox = ? voy = 0 (horizontal projectile)

ax = 0 ay = -9.81 m/s2

Step 2: Write down equations that might solve it.

g

yt

2

• We can use the y equation to solve for the time of the jump.

• We then use the time in the x equation to solve for the initial velocity.

tvx ox

Step 3: Carry out the solution. Time for the jump

• Using the y equation to solve for the time of the jump.

g

yt

2

s 378.0 9.81-

) 2(-0.7

t

t

Step 3: Carry out the solution.Using our time to calculate the initial horizontal velocity

• Using the x equation to solve for the initial velocity.

m/s 0.66

s 378.0

m 25.0

ox

ox

vt

x

tvx

Let us review the concepts for Horizontally launched projectiles

• Acceleration due to gravity does not affect horizontal motion

• Initial horizontal velocity = final horizontal velocity•

• Initial vertical velocity = 0m/s•

3.3 Projectile Motion

The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. a) Determine the time required for the Care package to hit the ground.b) What are the magnitude and direction of the final velocity of the Care

package?

Practice Problem # 2 - A Falling Care Package

Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)

X- Component Y - Component

Dx = Dy = -1050.0 m (negative due to downward direction of fall)

vox = 115 m/s voy = 0 (horizontal projectile)

ax = 0 ay = -9.81 m/s2

Step 2: Write down equations that might solve it.

• We can use the y equation to solve for the time of the jump.• Since

g

yt

tgy

tgtvy oy

2

))((

))((2

21

221

Step 3: Carry out the solution.Time for the jump

• Therefore, using the modified y equation to solve for the time of the jump.

g

yt

2

s 6.14 9.81-

) 2(-1050

t

t

3.3 Projectile Motion

b) What are the magnitude and direction of the final velocity of the Care package?

Practice Problem # 2 - A Falling Care Package

Practice Problem # 1Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table)

X- Component Y - Component

Dx = Dy = -1050.0 m (negative due to downward direction of fall)

vox = 115 m/s voy = 0 (horizontal projectile)

ay = -9.81 m/s2

t = 14.6 s t = 14.6 s

Step 2: Write down equations that might solve it.

• We then use the time in the y equation to solve for the final vertical velocity.• Since

tavv yoyy

3.3 Projectile Motion

s 6.14sm80.90 2

tavv yoyy

sm143

Step 3: Carry out the solution.Final vertical velocity of the package

Step 4: Carry out the solution.Magnitude and direction of final velocity

22yx vvv

)143(11522

v

smv /184

115

143tan 1

2.51

CLASS ASSIGNMENT NOT H/W

I will not accept late work

3.3 Projectile Motion

Conceptual Example 5 I Shot a Bullet into the Air...

Suppose you are driving a convertible with the top down.The car is moving to the right at constant velocity. You pointa rifle straight up into the air and fire it. In the absence of airresistance, where would the bullet land – behind you, aheadof you, or in the barrel of the rifle?Explain

Section 3. Cont.

Projectiles at an Angle

Projectiles Launched at an angle

As shown below, projectiles launched at an angle have an initial vertical and horizontal component of velocity.

• Suppose the initial velocity vector makes an angle θ with the horizontal • The sine and cosine functions can be used to find the horizontal

and vertical components of the initial velocity

Projectiles Launched at an angle

3.3 Projectile Motion

Example 6 The Kickoff

• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:

a) determine the maximum height that the ball attains.

Step 1: Write down what you know and don’t know (and resolve the vector provided)

X – Component Y - Component

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

Dy = ?

Dt = ?Dt = ?

)cos(oox vv )sin(ooy vv

Dx = ?

=22 m/s

=40o

Step 2: Decide on a plan or equation to use.

• For the horizontal velocity:

• For the vertical velocity:

sm1.1440sinsm22sin ooy vv

sm9.1640cossm22cos oox vv

Step 3: Carry out the plan.

y ay vy voy t? -9.81 m/s2 0 14.1 m/s

yavv yoyy 222 y

oyy

a

vvy

2

22

m 1.10

sm81.92

sm1.1402

2

y

3.3 Projectile Motion

Example 6 The Kickoff

• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:

a) The maximum height that the ball attains = 10 mb) The time between kickoff and landing

Step 1: Write down what you know and don’t know (and resolve the vector provided)

X – Component Y - Component

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

Dx = ? Dy = 10.1 m

Dt = ? Dt = ?

smvox /9.16 smvoy /1.14

40ovoy

vox

V i = 22 m

/s

Step 2: Decide on a plan or equation to use.

• We can calculate tine for the vertical displacement:

tavv yoyy

• For the vertical displacement (to maximum height):

• = 2.87 s

tavv yoyy

a

vvt oyy

Step 3: Carry out the plan.

2)81.9

1.140( xt

81.9

1.141.14

t

3.3 Projectile Motion

Example 6 The Kickoff

• A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following:

a) The maximum height that the ball attains = 10 mb) The time between kickoff and landing = 2.86 sc) The range ‘R’ of the projectile?

Step 1: Write down what you know and don’t know (and resolve the vector provided)

X – Component Y - Component

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

Dx = ? Dy = 10 m

Dt = 2.86 s Dt = 1.43 s

smvox /9.16 smvoy /1.14

40ovoy

vox

V i = 22 m

/s

Step 2: Decide on a plan or equation to use.

• For the horizontal displacement:

tvtatvx oxxox 221

• For the horizontal displacement:

• = 48.5 m

Step 3: Carry out the plan.

ssm 87.2/9.16

tvtatvx oxxox 221