chapter 7 projectile motion kinematics in dimensions two

Chapter 7 Projectile motion

Kinematics in Dimensions

Projectile Motion

• Projectile motion is motion in two directions

• Motion in the x-direction is independent of the y-direction

• Motion in the y-direction is independent of the x-direction

What is a Projectile?

• A projectile is any object that is placed into free flight and is being affected by gravity.

• The path of a projectile is called the trajectory.

28.9s

mg

• The trajectory of a projectile forms the shape of a parabola.

More on Trajectories

Different Trajectories

• Depending on initial location and angle, projectiles can form different paths, all of which possess some parabolic shape.

This path is called a “half” parabola.

The Half Parabola

Trajectory

• An object is launched horizontally and allowed to fall.

• There is no vertical velocity at the start, v1y is zero.

• The time of flight is equal to the time it would take to drop the object from rest.


• Depending on initial location and angle, projectiles can form different paths, all of which still possess some parabolic shape.

This path is called a “partial” parabola.


• Depending on initial location and angle, projectiles can form different paths, all of which still possess some parabolic shape.

This path is called a “full” parabola.

Throw

The Full Parabola

Vy Vx

• The key to the full parabola is symmetry.

• Try to identify some points of symmetry.

Quick Comparison of Paths

• Half Parabola– Launched Horizontally from

some height

• Full Parabola– Launched at an angle from

ground level– Symmetrical trajectory

• Partial Parabola– Launched at an angle from some

height

Half Parabola Timing• The time of flight of a half parabolic

path is equal to that of simply dropping the object from the same height.

Horizontal velocity (vx) has no affect on flight time because it

is not affected by gravity.

Jill drops the yellow ball off of a cliff. What happens to the ball?

Does it have constant velocity?

Now Jill drops the yellow ball and throws the red ball horizontally. Which ball will hit the ground first?

Half Parabola Summary• Objects must be dropped from some height d1y.

• The vertical reference point is usually the ground or floor.

• The time of projectile’s flight is identical for that of simply dropping the object the same distance (straight down).

• Horizontal velocity remains constant in all projectile problems (v1x = v2x).

212 1 1 2yy y v t gt

2xd1xd

1yd

2 yd

http://www.cs.sbcc.cc.ca.us/~physics/flash/projectilemotion.html

A stone is thrown horizontally from the top of a 78.4m high cliff at 5m/s.

a) How long does it take to reach the bottom?

2 22 1

2 1

2v v gd

v v gt

21

1 2

1

2( )

d v t gt

d v v t

Var X Y Want

a

v1

v2

d1

d2

t



2yt

g

2y y

1d = v t + at

2

2

2( 78.4 )

9.8ms

mt

4.0t s


b) How far from the base does it land?

x xd = v t

5 (4.0 )msxd s

20xd mxd


c) What are the final vy and vx ?

2 1x xv v

22 1 2y yv v ad

22

2 (0 ) 2( 9.8 )( 78.4 )m msy s

v m

2 39.2 /yv m s

2 22 1 2y yv v ad

Sample Partial• A cannon nestled in the side of a cliff (d1y = 65m)

fires a cannon ball at 26 . How long until the ball splashes into the sea?

Fire

ms

Sample Problem

• A toy car is raced off a table (1.1m high) onto the floor below.– How long did it take for the car to crash on the floor?

2 12 y yd dt

g

212 1 1 2y y yd d v t gt

2

2 0 1.1

9.8 ms

m mt

0.474t s

0

Object launched horizontallyMulti media studio

• Horizontally launched projectiles

• The plane and the package

• The truck and the ball

What do you notice about the horizontal velocity in each of the following animations?

Projectile Motion Type

Not all object are launch horizontally

Objects can be launched at an angle

• Recall the trajectory of the golf ball when hit with a 3 iron.

• What would the trajectory of a 9 iron look like?• The loft of the club changed the launch angle.

Projectile Motion Path Plot

0

5

10

15

20

25

30

0 10 20 30 40 50 60

Horizontal Position dx (m)

Ve

rtic

al

Po

sit

ion

dy

(m)

• Object 1 was launched at 60o


• Object 1 was launched from a 25m high cliff at 0o


Projectile Motion Path Plot

0

5

10

15

20

25

30

0 10 20 30 40 50 60

Horizontal Position dx (m)

Ve

rtic

al

Po

sit

ion

dy

(m)

Projectile motion generator

Analytical Vector Addition• You will need to understand the basic trig

functions.

sinopp

hyp

cosadj

hyp

tanopp

adj

opp

adj

hyp

Initial Velocity Breakdown• When an object is launched at some angle, it’s initial

velocity (v1) can be broken down into two components.

– Horizontal Component (Vx)

– Vertical Component (Vy)

• What shape is formed?• Consider also the launch angle ().

v1v1y

v1x

Please Note: horizontal and vertical components are independent of one

another. The only commonality is time.

Initial Velocity Breakdown (Cont.)• Consider the breakdown from the previous slide again. • There are trigonometric relationships between the sides

and angles of a right triangle.

tanopp

adj 1

1

tan y

x

v

v

sinopp

hyp 1

1

sin yv

v

cosadj

hyp 1

1

cos xv

v

v1 v1y

v1x

Important!

Practicing Trig Functions• Consider the triangle below.

• Solve for the unknown values.

13

x

y

22.62°

sinopp

hyp cos

adj

hyp

sin 22.6213

y

13 sin 22.62y

5y

cos 22.6213

x

13 cos 22.62x

12x

Searching for x Searching for y

Sample Velocity Breakdown• A dart gun is fired at an angle of 30° with a

muzzle velocity of 40m/s.

• Calculate the components of the velocity?

v1v1y

v1x

1

1

cos xv

v

1 1 cosxv v

1

1

sin yv

v

1 1 sinyv v

1 40 cos 30mx sv 1 40 sin 30my sv

1 34.6 mx sv 1 20 my sv

Horizontal Component (x)Vertical Component (y)

1 34.6 mx sv

1 20 my sv

Object launched at an angleMulti media studio

• Parabolic motion of projectiles

• Non-Horizontally launched projectiles

• Maximum Range

• Monkey and the Zoo Keeper

• Given an initial velocity of 40m/s and an angle of 25

• find v1x & v1y

sinO

H cos

A

H

sin yv

v cos xv

v

V 1=40

v1y

v1x

=25o

Searching for x

40cos 25oxv

36.25xv

Searching for y

40sin 25oyv

16.9yv

Sample Full Parabola Problem• A golf ball is struck at an angle of = 36° with

the horizontal at a velocity of 45m/s.– What are the components of the velocity (v1x and v1y)?

Strike

v1x

v1y

v1

1 1 cosxv v 1 45 cos 36m

x sv

1 1 sinyv v

1 45 sin 36my sv

1 36.4 mx sv

1 26.4 my sv

Horizontal Component

Vertical Component

Practice Problems

Homework• WS 7b 1-2• WS 7c 1-3

X & Y are Independent

Problem Solving Strategies

• Solve for the horizontal component Vxi

– Use trig functions

• Solve for the vertical component Vyi

– Use trig functions

• Solve each direction (x & y) separately

• Symmetry can be used when the launching & landing places are the same height.

a) Find the hang time

b) find the horizontal distance the ball travels.

b) The maximum height of the ball.

A football player kicks a ball at 27m/s at an angle of 30°.

Horizontal VerticalV K W V K W

a g -9.8

V1x v1y

v2x v2y

dx dx dy dy

t t t

Problem Solving StrategiesStep 1: Solve for the horizontal and vertical components

(V1x & V1y

)

V=27m/s

V1x= ?m/s

V1

y= ?

m/s

23.4 mx sv

27sin 30yv 27cos30oxv

13.5 my sv

Searching for x

cosxv v

30o

Searching for y

sinyv v

Problem Solving StrategiesSymmetry can be used when the launching & landing places are the same height.

12.5m/s

10.0m/s

7.50m/s

5.00m/s

0.00m/s

15.0m/s

2.50m/s


b) find the horizontal distance the ball travels.

b) The maximum height of the ball.



a g -9.8

v1x 23.4 V1y 13.5

v2x 23.4 v2y

dx dx dy dy

t t t

2 1y yv vt

g

2.75t s

2 1v v gt



2 1v v

1 1( )v v gt

2

13.5 13.5

9.8

m ms s

ms

t

(23.4 )*(2.75 )mx sd s

64.3xd m

xx

dv

t

b) Find the horizontal distance


12.5m/s

10.0m/s

7.50m/s

5.00m/s

0.00m/s

15.0m/s

2.50m/s

A football player kicks a ball at 27m/s at an angle of 30 °.

c) Find the maximum height

What is true about the vertical velocity at the maximum height?

A football player kicks a ball at 27m/s at an angle of 30°. Find the max height

2 22 1 2v v gd

9.3d m

2 22 1

2

v vd

g

2

2 20 (13.5 )

2( 9.8 )

ms

ms

d

An arrow is shot at 44m/s at an angle of 60°• Find the maximum height of the arrow.• Find the horizontal distance the arrows

travels.• Find the hang time

@ (max) 0 my y sd v

Xd2 1@ 0y y yd v v

Problem Solving StrategiesStep 1: Solve for the horizontal and vertical components

(V1x & V1y

)

V=44m/s

V1x= ?m/s

V1

y= ?

m/s

22 mx sv

44sin 60yv 44cos 60oxv

38.1 my sv

Searching for x

cos xv

v

60o

Searching for y

sin yv

v


b) find the horizontal distance the arrows travels.

c) The maximum height of the arrow.

An arrow is shot at 44m/s at an angle of 60°


a g -9.8

v1x 22 v1y 38.1

v2x v2y

dx dx dy dy

t t t

2 1y yv vt

g

7.78t s

2 1y yv v gt



2 1y yv v

1 1( )y yv v gt

2

38.1 38.1

9.8

m ms s

ms

t

(22 / )*(7.78 )xd m s s

171.2xd m

xx

dv

t

b) Find the horizontal distance


A football player kicks a ball at 44m/s at an angle of 60°. Find the max height

2 22 1 2y yv v ad

74.0d m

2 22 1

2y yv v

da

2

2 20 (38.1 )

2( 9.8 )

ms

ms

d

Recall, vy=0 at dy max

Practice Problems

• WS 7c– 4

• WS 7d1-3

Concept Questions

• A stone is thrown horizontally from a cliff.• How would the x distance change if the stone

was thrown twice as fast?

x xd = v t x xd = (2v ) t

• dx distance would double

Concept Questions

• How would the v2y change if the stone was thrown twice as fast?

• Vx and Vy are independent

• Since it was thrown horizontally, it would not change.

Concept Questions

• How would the v2y change if the cliff was twice as high?

2 22 1 2y yv v gd 2 2yv gd

2 2 (2 )yv g d

2 2 2yv gd

The Partial Parabola• Recall, this path has elevation and launch angle.• The trajectory again has an apex.• This is mathematically the most complex path.

Fire

Sample Partial Parabola Problem• A cannon nestled in the side of a cliff (d1y = 20m)

fires a cannonball at 130m/s at a 30° angle.

Fire

Find v1x &v1y

Find y maxFind t1 to apexFind t2 apex to ground (half parabola)Find total timeFind x max

Sample Partial Parabola Problem• A cannon nestled in the side of a cliff (d1y = 20m)

fires a cannonball at 130m/s at a 40° angle.

Fire

• You are trying to win a prize by throwing an apple into a basket on top of a pedestal.

• The apple leaves your hand 1.00 m beneath the top of the pedestal.

• The apple flies 3.09 m horizontally before landing in the bottom of the basket.

• The apple’s maximum height was 1.26 m.• What was the apple’s initial velocity

(magnitude and direction)?

Partial ParabolaVariable x y

a

v1

v2

d1

d2

t

Partial Parabola• Find V1y knowing Dy max = 1.26m

Dy (max) = 1.26m

2 22 1 2y yv v gd

2 22 12y yv gd v

22 12y yv gd v

2

21 (0 ) 2( 9.8 )(1.26 )m my s sv m

1 4.97 my sv

Partial ParabolaFind the hang time knowing v1y and ending height d2y=1m

22 1

1

2yd v t gt

2 4

2

b b acx

a

1

1

2

y

a g

b v

c d

.73t s

21 1 4(.5)( )( )

2(.5)( )y yv v g d

tg

Partial ParabolaPartial parabolas can be represented by two half parabolas

Total hang time is the time of the ½ parabola on the way up plus the time of the ½ parabola on the way down.

2 1y yv vt

g

2 ydtg

.5t s .23t s

.73t s

The Partial Parabola• If you look at this path carefully, you can see two

half parabolas, which simplifies things considerably.

• You still must consider the launch angle and the components of the velocity when trying to solve.

Practice Problems

• WS 7c1-6

8-9

Mortar Problems • A mortar crew fires a projectile at

an enemy ammunitions storage facility that is protected by a wall located on top of a 200.0 m high cliff.

• The ammunition is located a horizontal distance of 314.68 m from the mortar’s position.

• The projectile passes directly over the wall at its maximum height of 215.24 m.

• What was the projectile’s initial velocity (magnitude and direction)?

V x y

a

v1

v2

d1

d2

t

Mortar Problems

EndChapter 7 Projectile motion

Kinematics in DimensionsTwo

Simple Harmonic Motion

Simple Harmonic Motion: Motion caused by a linear restoring force that has a period independent of amplitude.

Period: The time required to repeat one complete cycle

Amplitude: Maximum displacement from equilibrium.

Restoring force

Equilibrium Position

Periodic Motion is a form of Simple Harmonic Motion

Restoring force

Simple Harmonic Motion

Conservation of Energy

A stone is thrown horizontally from the top of a 44m high cliff at 15m/s.


Var Known Want

a 9.8m/s2

vyi

vyf

d1 44m

d2 0m

t t

2 22 1

2 1

2v v ad

v v at

21

1 2

1

2( )

d v t at

d v v t



2yt

g2

y y

1d = v t + at

2

2

2( 44 )

9.8ms

mt

3.0t s


b) How far from the base does it land?

x xd = v t

(15 )*(3.0 )mx sd s

45xd mxd

Acceleration due to Gravity

Free FallIf you drop a book and a piece of paper which will hit the floor first?

Acceleration due to GravitySimply replace “a” with “g”.

21

1

2d v t gt

2 1v v at

1 2

1

2d v v t

21

1

2d v t at

2 22 1 2v v ad

2 1v v gt

1 2

1

2d v v t

2 22 1 2v v gd

Acceleration due to GravityGravity Time Instant

speedAverage

speed

Distance

10m/s2 0s 0m/s 0m/s 0m

10m/s2 1s 10m/s 5m/s 5m

10m/s2 2s 20m/s 10m/s 20m

10m/s2 3s 30m/s 15m/s 45m

10m/s2 4s 40m/s 20m/s 80m

10m/s2 5s 50m/s 25m/s ½ gt2

What geometric shape would that object have on a position versus time graph after experiencing acceleration?

Position - Time Graph

0

100

200

300

400

500

600

0 1 2 3 4 5 6 7 8 9 10

Time (s)

Posit

ion

(m)

Graph shapes

• Given an initial velocity of 30m/s and an angle of 35

• find v1x & v1y

sinO

H cos

A

H

sin yv

v cos xv

v

V 1=30m/s

v1y

v1x

=35o

Searching for x

30 cos35omx sv

23.9 mx sv

Searching for y

30 sin 35omy sv

18.0 my sv

chapter 7 projectile motion kinematics in dimensions two

Documents

height slide

xdirection slide

half parabola trajectory

half parabola timing

projectile motion kinematics

ydirection motion

time of flight

flight time