Two-Dimensional MotionChapter 3

A little vocab

Projectile = any object that moves through space acted on only by gravity

Trajectory = the path followed by a projectile

Range = the horizontal distance a projectile travels

Puzzle

If a ball is dropped straight down at the same time one is launched horizontally, which ball will hit the ground first?

Does this work the same if it is a bullet launched horizontally, at the same time a ball is dropped from the same height?

Without any means to propel an object, once it is moving horizontally, it has constant horizontal motion… x = vt

(no acceleration)

In free fall, vertical motion… y = vit + ½ gt2

We know…

Put these motions together

Motion in Two Dimensions

Using + or – signs is not always sufficient to fully describe motion in more than one dimension Vectors can be used to more fully

describe motion

Still interested in displacement, velocity, and acceleration

Projectile Motion

An object may move in both the x and y directions simultaneously It moves in “two dimensions”

The form of two dimensional motion we will deal with is an important special case called projectile motion

Assumptions of Projectile Motion

Because this is a “perfect physics world” We may ignore air resistance and friction We may ignore the rotation of the earth

With these assumptions, an object in projectile motion will follow a parabolic path

Rules of Projectile Motion The x- and y-directions of motion are

completely independent of each other X-motion and Y-motion happen at the

same time ay = g

Why? The y-direction is free fall

ax = 0 Why? The x-direction is uniform motion There is nothing else pushing in the horizontal

direction

Horizontal Projectile Motion

something thrown horizontally Without gravity, it would continue along a

horizontal path. Because gravity acts “downward”, the path is half

of a parabola.

• Start at maximum height

• Given a push in the X-direction only

Solving Horizontal Projectile Motion Make a chart with 2 sides What do we ALREADY know (without even

having a problem to solve?)Horizontal Vertical

a =

vi =

vf =

t =

x =

a =

vi =

vf =

t =

x =

0

#

# = vi

#

# = range

9.8m/s2 = g

0

# (just before it hits)

#

Max height = y

NOTE: Can only cross over at t!

Solving contunued…

Use kinematics equations to solve Horizontal… if a = 0, the only

equation that matters is #2 x = vit

Vertical …1. vf = vi + gt

2. y = vit + ½gt2

3. vf2 = vi

2 + 2gy

HINT: Solve for time first… goes on both sides of chart!

Example

A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 32 cm below the height from which it was thrown. How far away is the player from the board?

Horizontal Vertical

a =

vi =

vf =

t =

x =

a =

vi =

vf =

t =

y =

0

12.4 m/s

12.4 m/s

?

?

9.8m/s2

0

?

?

32cm = .32m

What can we solve first? We are looking for “x” Not enough info on the horizontal

side… On the vertical side, we can solve for

vf

t Solve for t, so we can use it on the

horizontal side Don’t really care about vf this time

The math…

For t

y = vit + ½gt2

.32 = 0 + ½ (9.8)t2

t = 0.2556s

Horizontal Vertical

a =

vi =

vf =

t =

x =

a =

vi =

vf =

t =

y =

0

12.4 m/s

12.4 m/s

0.2556s

?

9.8m/s2

0

?

0.2556s

32cm = .32m

For x

x = vt

x = (12.4)(.2556)

x = 3.17m

What if… the previous question asked for the

total final velocity? This is where VECTORS come in… We knew that vf was the same as the

vi in the horizontal = 12.4m/s We can find vf in the vertical

vf = vi + gtvf = 0 + (9.8)(0.2556)vf = 2.5m/s

We aren’t done yet.

Total velocity is a vector in polar coordinates.

12.4m/s

2.5

m/sResultant

(r,Θ)

r2 = 12.42 + 2.52

r = 12.6m/s

tanΘ = -2.5/12.4

Θ = tan-1(-0.2016)

Θ = -11.4o

Θ = -11.4o

Θ