Kinematics- 2 Dimensions

Projectile Motion

What is Projectile?

Projectile -Any object which projected by some means and continues to move due to its own inertia (mass) in a curved path

Projectiles are objects in free fall that have an initial horizontal velocity.

Free fall means the only force acting is gravity

Initial horizontal velocity results in motion in two dimensions

Examples of projectiles include

Bullets after they leave the barrel of the gun

Baseballs after they are thrown, unless they are thrown straight up

Projectiles move in TWO dimensions

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

Horizontal

Vertical

Horizontal and vertical motions are independent.

The common factor is time.

Horizontal “Velocity” Component

NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity

In other words, the horizontal

velocity is CONSTANT. (Vx = Vox)

BUT WHY?

Gravity DOES NOT work

horizontally to increase or

decrease the velocity. (ax = 0 m/s2)

Vertical “Velocity” Component

Changes (due to gravity), does NOT cover equal displacements in equal time periods.

Both the MAGNITUDE and DIRECTION change.

As the projectile moves up the MAGNITUDE

DECREASES & its direction is UPWARD. As it

moves down the MAGNITUDE INCREASES and

the direction is DOWNWARD.

Combining the Horizontal & Vertical Components

Together, these components produce what is called a trajectory or flying path. This path is parabolic in nature.

Component Magnitude Direction

Horizontal Constant Constant

Vertical Changes Changes

Horizontally Launched Projectiles

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

0 /oyv m s

constantox xv v

Horizontally Launched Projectiles

To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And we use the following kinematic equation: 21

2oxx v t at

oxx v t

Remember, the velocity

is CONSTANT

horizontally, so that

means the acceleration is

ZERO!

212

y gt

Remember that since the

projectile is launched

horizontally, the INITIAL

VERTICAL VELOCITY is

equal to ZERO.

Horizontally Launched Projectiles

• Example: A plane traveling with a horiz-ontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?

What do I

know?

What I want to

know?

vox=100 m/s t = ?

y = 500 m x = ?

voy= 0 m/s

g = -9.8 m/s2

2 2

2

1 1500 ( 9.8)2 2

102.04

y gt t

t t

10.1 seconds

(100)(10.1)oxx v t 1,010 m

Vertically Launched Projectiles

Component Magnitude Direction

Horizontal Constant Constant

Vertical Decreases up, 0 @

top, Increases down Changes

Horizontal Velocity

is constant

Vertical

Velocity

decreases

on the way

upward

Vertical Velocity

increases on the

way down,

NO Vertical Velocity at the top of the trajectory.

Vertically Launched Projectiles

Since the projectile was launched at a angle, the velocity MUST be broken into components!!!

cos

sin

ox o

oy o

v v

v v

vo

vox

voy

Vertically Launched Projectiles

There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0

Vertically Launched Projectiles

You will still use kinematic equation, but YOU MUST use COMPONENTS in the equation.

cos

sin

ox o

oy o

v v

v v

vo

vox

voy

oxx v t 212oyy v t gt

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?

(b) How far away does it land?

(c) How high does it travel?

cos

20cos53 12.04 /

sin

20sin 53 15.97 /

ox o

ox

oy o

oy

v v

v m s

v v

v m s

53

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?

What I know What I want to

know

vox=12.04 m/s t = ?

voy=15.97 m/s x = ?

y = 0 ymax=?

g = - 9.8 m/s2

2 2

2

1 0 (15.97) 4.92

15.97 4.9 15.97 4.9

oyy v t gt t t

t t t

t

3.26 s

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(b) How far away does it land?

What I know What I want

to know

vox=12.04 m/s t = 3.26 s

voy=15.97 m/s x = ?

y = 0 ymax=?

g = - 9.8 m/s2

(12.04)(3.26)oxx v t 39.24 m

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(c) How high does it travel?

CUT YOUR TIME IN HALF!

What I know What I want

to know

vox=12.04 m/s t = 3.26 s

voy=15.97 m/s x = 39.24 m

y = 0 ymax=?

g = - 9.8 m/s2

2

2

12

(15.97)(1.63) 4.9(1.63)

oyy v t gt

y

y

13.01 m

X Motion-

Constant velocity

Y-Motion-

Constant Acceleration

tvx x

2

21 tatvy yoy

Equations of Motion for projectiles

2m/s 81.9ya

constant xfxox vvv

tavv yoyfy

0xa

yavv yoyfy 222

A scared kangaroo once cleared a fence by jumping with a speed of 8.42 m/s at an angle of 55.2o from the ground. If the jump lasted 1.40 s, how high was the fence? What was the kangaroo’s horizontal displacement?

Hint: The time for the maximum height was half the total time.

Kangaroo Problem

Step 1: Write down givens, unknowns and model.

x = ?

55.2o

t = 1.40s for

whole jump.

t= 0.70 s for

maximum

height.

Projectile launched at an angle

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

tvx x

For the horizontal

displacement:

For the vertical displacement:

2

21 tatvy yiy

2s

m81.9yaWe will use trigonometry to find the initial

velocities.

Note: The only triangle here is initial velocity and its components!

Step 3: Carry out the plan.

)cos(i

voxv

)cos(i

voxv

sm91.6

)2.55sin()sm42.8(

)sin(

)sin(

oyv

oyv

ivoyv

iviy

v

o

)2.55cos( ) 42.8(sm ooxv

vox

voy

sm81.4oxv

Step 3: Carry out the plan.

m 6.7s) 4.1)( 81.4(sm x

tvx xs 1.4 t 81.4 s

m xox vv

2

21 attvy oy

2s

m m/s 81.9 s 0.70 t 91.6 avoy

m 43.2403.2837.4

)s 7.0)(81.9(- s) 7.0)(91.6( 2

s

m21

sm

2

y

y

Step 4: Check your answer.

x = 6.7 m

55.2o

t = 1.40s

Baseball Problem

A baseball is thrown at an angle of 25o relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?

Baseball Problem

m/s 23.0 m 042 i v. x We will split the initial

velocity into x and y components.

We can use the x-equation to find the time it was in the air.

We can then use the y-equations to find the maximum height.

θ=25o

This is a projectile. The

x motion is constant

velocity. The y motion is

constant acceleration.

twhole flight =?

ΔyMAX = ?

Baseball Problem

m/s 23.0 m 042 i v. x

t vx x tv

x

x

m/s 20.8

)25)(cos(023()25cos(sm

oxx

iox

vv

. v v oo

s 0.2 20.8

m 42

sm

t

vox

m/s 9.72

)25)(sin(023()25sin(sm

oo . v v ioy

vox

Baseball Problem

The highest point will be half-way through the flight. t = ½ x 2.0 s = 1.0 s

m 8.4

s) 0.1)(m/s 81.9( - s) m/s)(1.0 72.9( 22

21

y

y

2

21 att vΔy oy