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Chapter 4

Two-Dimensional

Kinematics

Units of Chapter 4

• Motion in Two Dimensions

• Projectile Motion: Basic Equations

• Zero Launch Angle

• General Launch Angle

• Projectile Motion: Key Characteristics

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4-1 Motion in Two Dimensions

If velocity is constant, motion is along a straight line:

If velocity is not constant, there is acceleration.

When acceleration and velocity are along the same

line, motion is still along a straight line. (Chapter 2)

What if velocity and acceleration are not along the

same line? Motion will no longer be in 1D.

4-1 Motion in Two Dimensions

V3 = V1+ V2

V1

V2

V3

V1

V2

V3

Consider one motion has two parts of velocity. If V1 and V2 are not perpendicular to each other, they are related.

V2 has component along V1’s direction

V1 has component along V2’s direction

If V1V2, one East, one North, they never contribute to each other!

V2

V1

Motions in two perpendicular directions

are INDEPENDENT from each other.

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4-1 Motion in Two Dimensions

Motion in the x-and y-directions should

be solved separately:

1.) Define Coordinate System

x direction: horizontal

y direction: (positive can be up or down)

4-2 Projectile Motion: Basic Equations

Assumptions:

• ignore air resistance

• g = 9.81 m/s2, downward

• ignore Earth’s rotation

x-direction is a horizontal direction.

acceleration in x-direction is zero

If y-axis points upward,

acceleration in y-direction is -9.81 m/s2

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4-2 Projectile Motion: Basic Equations

The acceleration is independent of the direction

of the velocity:

4-2 Projectile Motion: Basic Equations

These, then, are the basic equations of

projectile motion:

(if up is defined to be positive)

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4-3 Zero Launch Angle Launch angle:

direction of initial velocity with respect to horizontal

When the lunch angle is zero,

the initial velocity in the

y-direction is zero.

Initial velocity is in the

x-direction v0x=v0

Example 1:

A

x

30m

B vx0 t

vx0

vy 0

x

Question: 1.)Which one touches the ground first?

(what if Ball B was thrown upward or downward?) Answer: 1.) They touch ground at the Same time.

Because the motion in y direction (vertical) is

INDEPENDENT to motion in x direction (horizontal). Question: 2.)For B from cliff to water level, what are the known items?

~Ball A falls from rest

~Ball B was shot out

horizontally at v0= 10 m/s

Define right to be +x

Define up to be +y

y

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Question: 3.) How long it takes for them to hit ground?

Hint: Which direction’s motion is limiting the traveling

time before it reach the ground?

Answer: 3.) The y direction. Because from roof to the

ground, y changes. The magnitude of y determines

the total time to fall. To find time, we use the equation without vfy

y = v0yt + ½ ayt2

v0y= 0 ;

t2 = 2y/ay

Don’t memorize these conclusions. The don’t work for

different problems. LEARN METHORDS NOT RESULTS.

Question: 4.)Where is Ball B when it hits the ground?

Hint: Which direction’s motion should you use to

decide Ball B’s position when it hits the ground?

Answer: 4.) Use x direction! We know y= 0 at the

end. We need to find x at the end.

x = v0xt = 10(m/s) x 2.47s = 24.7 m

Don’t worry about x = v0xt + 1/2axt2,

because ax= 0 !!!

x = v0xt

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Question: 5.)What is the final velocity of Ball B when

hits the ground? Find both direction and speed?

vfx= v0x= 10m/s vx didn’t change!

vfy= v0y+ ayt= 0 –9.8 x 2.47 = -24.2 m/s

v = (vx2+ vy

2) = (102+24.22) = …m/s

q= tan-1 (vy/vx)

= tan-1(-24.2/10)

= -tan-1(24.2/10)=-67.5o

q

4-4 General Launch Angle

Treat x(horizontal) and y(vertical) directions separately.

When angle is not zero, v0x = v0 cos θ

v0y = v0 sin θ

If up is + y direction:

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Example 2: v0 = 5.00 m/s, q= 30º, x0=0, y0=0

Question: 1) Find total time of flight before landing

on the ground? Which final point? Which

direction? Which equation?

Y direction limits travel time! y = v0y t - ½ gt2 = 0 v0y t= ½ gt2

so t=2 v0y /g = 5.0/9.81= 0.51(s) (Don’t memorize)

v0x= v0cos 30º= +4.33 m/s

v0y= v0sin 30º= +2.50 m/s

Example 2: v0 = 5 m/s, q= 30º, x0=0, y0=0

v0x= v0cos 30º= +4.33 m/s

v0y= v0sin 30º= +2.50 m/s

Question: 2) Distance of landing? Which final point?

Consider Which direction? Which equation?

X direction motion determines landing distance x= v0x t Landing total time t= 0.51 (s)

x = 4.33 ×0.51= 2.2 (m) (Don’t memorize)

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Example 2: v0 = 5 m/s, q= 30º, x0=0, y0=0

Question: 3) Find

Final velocity before

landing on the

ground, both

direction and speed.

Consider both x and

y directions.

vx=v0x= 4.33 (m/s) constant.

vy= vy0 -gt= 2.5 –9.81×0.51= -2.5 (m/s)

Same magnitude as vy0. Opposite direction.

Final velocity: v = (vx2+ vy

2) = 5 m/s

q= tan-1 (vy/ vx) = tan-1 (-2.5/ 4.33) = -30

^ v = (4.33 m/s)x + (-2.5 m/s)y ^

t=0.51 s

^ v0 =(4.33 m/s)x + (2.5 m/s)y ^

Example 2: v0 = 5 m/s, q= 30º, x0=0, y0=0

Notice the Symmetry of

Projectile Motion:

If it lands at the same height

t of first (raising) half

=

t of second (falling) half.

Why?(read it yourself)

y = v0yt + ½ at2= (v0 + at) × t - ½ at2

y = vfyt - ½ at2 Page 100, CE, 5 & 6

Raising vfy= 0, a = -g; vy=0-v0y; traise= v0y/g Falling v0y= 0, a = -g; vy=vfy-0=-v0y tfall= v0y/g

yfall= v0ytfall+ ½ atfall2= - ½ gtfall

2

yfall= -yraise ytotal= 0 traise=tfall= (total time)/2

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x for the first half

=

x for the second half

vx0×traise=vx0×tfall

Vfy is opposite to v0y

because Vfy =vtop+ atfall= 0 –gtfall

vfy=-v0y vtop=v0y+ atraise= 0

v0y= -atraise= gtraise

Final speed, |vf|=(vfx2+ vfy

2) = (v0x2+ v0y

2) = v0

Final velocity has same magnitude (speed) as

launch velocity, angle q, becomes -q.

Example 2: v0 = 5 m/s, q= 30º, x0=0, y0=0

v0x = v0cos30º= +4.33 m/s

v0y = v0sin30º= +2.50 m/s

Question: 4) Find maximum height it can reach. Which

final point? Which direction? Which equation?

Y direction determines maximum height

vyf2= v0y

2+ 2ay vyf2= v0y

2-2gy

y =( vyf2-v0y

2)/2g= -2.52/(-2 ×9.81)= 0.32 m

5) Time to reach maximum height? vfy= v0y - gtraise=0 traise= v0y/g

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4-4 General Launch Angle

Snapshots of a trajectory; red dots

are at t = 1 s, t = 2 s, and t = 3 s

The path followed by a projectile is a parabola.

4-5 Projectile Motion: Key Characteristics

Range: the horizontal distance a projectile travels

If the initial and final elevation are the same:

Do not memorize.

x = v0x t = v0x ×2(v0y /g)

= v0cosq ×2v0sinq/g

=v02 (2cosq sinq) /g

=v02 (sin2q)/ g

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4-5 Projectile Motion: Key Characteristics

The range is a maximum when θ = 45°, 2θ = 90°

Range= v02 (sin2q)/ g

Summary of Chapter 4

• Components of motion in the x-direction and y-

directions can be treated independently

• In projectile motion, the acceleration is

downward (g), in y direction.

• There is no acceleration in x-direction, (ignore

air resistance).

• If the launch angle is zero, the initial velocity

has only an x-component

• The path followed by a projectile is a parabola.

• The range is the horizontal distance the

projectile travels

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Projectile Motion: Problem solving:

Notice that y is not zero, if it lands at different

height.

1, Define x and y directions

2, Decompose initial v in x and y directions

3, Identify final point and label known and unknown.

4, Choose and solve equations.