How to do kinematics in 2-D Remember the ball shot at the monkey. Motion, force and accelerations in the X direction do not affect Y motion. And vice versa:

How to do kinematics in 2-D

• Remember the ball shot at the monkey. Motion, force and accelerations in the X direction do not affect Y motion.

• And vice versa: motion, force and accelerations in the Y direction do not affect X motion.

• So motion needs to be treated as two separate problems: 1 problem along the X axis and 1 problem in the Y axis and the two problems DO NOT AFFECT each other.

How to do kinematics in 2-D

• 1) Step 1: List x and y variables separately. Use geometry or trig to separate into components (use cos, sine, tangent). (or pick a new frame of reference to make the cosine and sine go away. Example, the escalator problem. Slant the angle of the X axis.)

• 2) Treat x and y like two different problems.

• 3) Solve x and y.

• 4) Recombine x and y vectors with geometry or trig to find the Resultant Vector. (Find the resultants or total vectors)

Vector Review, Displacement

• Δx = “horizontal displacement”

• Δy = “vertical displacement”=> “total displacement” = Magnitude and Angle

• vx = “horizontal component of velocity”

• vx = vtotal cosΘ

• vy = “vertical component of velocity”

• vy = vtotal sinΘ

• vtotal=

Velocity

Angle Review• Trig: For any triangle:

• From SOHCAHTOA

• Θdisp = cos-1 (X/H) Θvel = cos-1 (VX/Vtotal)

• Or

• Θ disp = tan-1 (Y/X) Θvel = tan-1 (VY/VX)

Acceleration

• ax = “horizontal component of acceleration”

• ax = atotal cosΘ

• ay = “vertical component of acceleration”

• ay = atotal sinΘ

• atotal=

• Find the angle also.

2-D Kinematic Equations

• Δx = xi + vixt+ ½ axt2

• vfx 2

= vix 2+ 2axΔx

• Δvx = axt

• Δy = yi + viyt+ ½ ayt2

• vfy 2

= viy 2+ 2ay Δy

• Δvy = ayt

Note: t is the only variable common to both dimensions.

2-D Kinematic Equations

• Δx = xi + vixt+ ½ axt2

• vfx 2

= vix 2+ 2axΔx

• Δvx = axt

• Δy = yi + viyt+ ½ ayt2

• vfy 2

= viy 2+ 2ay Δy

• Δvy = ayt

Freefall: This is a special case where ax = 0 ay = -9.80 m/s2 = GravityAssume: No air resistance, no friction, no forces other than gravity.

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.


Example 1 A Moving Spacecraft

In the x direction, the spacecraft has an initial velocity componentof +22 m/s and an acceleration of +24 m/s2. In the y direction, theanalogous quantities are +14 m/s and an acceleration of +12 m/s2.a)Total initial velocity and acceleration

Find, after 7.0 s.b) Δx and vfx, c) Δy and vfy, and d)total final velocity


ivsm14iyv

sm22ixv

sm26sm14sm22 22 iv

322214tan 1


iasm12iya

sm24ixa

sm27sm12sm24 22 ssai

272412tan 1

2

2

2


Example 1 A Moving Spacecraft

In the x direction, the spacecraft has an initial velocity componentof +22 m/s and an acceleration of +24 m/s2. In the y direction, theanalogous quantities are +14 m/s and an acceleration of +12 m/s2.Find (a) x and vx, (b) y and vy, and (c) the final velocity of thespacecraft at time 7.0 s.

x ax vx vox t? +24.0 m/s2 ? +22 m/s 7.0 s

y ay vy voy t? +12.0 m/s2 ? +14 m/s 7.0 s


x ax vx vox t? +24.0 m/s2 ? +22 m/s 7.0 s

m 740s 0.7sm24s 0.7sm22 2221

221

tatvx xox

sm190s 0.7sm24sm22 2

tavv xoxx


y ay vy voy t? +12.0 m/s2 ? +14 m/s 7.0 s

m 390s 0.7sm12s 0.7sm14 2221

221

tatvy yoy

sm98s 0.7sm12sm14 2

tavv yoyy


vsm98yv

sm190xv

sm210sm98sm190 22 v

2719098tan 1


Projectile motion

• Most common 2D kinematics type.

• An object gets “thrown” at some angle.

Projectile Motion (Type of Freefall)

• 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

• We make a few assumptions:– Object has no propulsion of its own.– The only acceleration on the object is gravity.– We ignore air friction– We ignore the rotation of the earth

(Question:

How do we ignore the earth’s rotation???)

Assumptions of Projectile Motion

– We may ignore the rotation of the earth

(How do we ignore the earth???)

ANS: Put our frame of reference on the surface of the earth so that the frame of reference rotates as the earth rotates. i.e., we use a moving origin, fixed to the surface of the earth.

Free fall problems, pg 1/2

• y-direction– viy = vitotal sinΘ

– Free fall problem• ay = -g = -9.80 m/s2

– Take the positive direction as upward– Uniformly accelerated motion, so the motion

equations all hold

Free fall problems, pg 2/2

• x-direction– vix = vitotal cosΘ

– No acceleration• ax = 0

• vix = vfx

• Δx = vix t

– Usually: • Right X is positive, left X is negative, or• East is positive, west is negative.

Free fall problems

• What does the path of a projectile (i.e. a launch object in freefall) look like?

• (Launch Monkey ball demo here at 3 different angles. Have class guess.)

• Answer on next slide.

The result

• A parabola

See, it’s a parabola

Time of Travel• Time of flight does not depend on vx or Δx at all.

• Time of flight is only determined by the Y motion.

No matter what angle you start at, it’s a parabola.

• The maximum range occurs at a projection angle of 45o

• Complementary values of the initial angle result in the same range– The heights will be

different

Example A Falling Care Package

The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time requiredfor the care package to hit the ground.

3.3 Projectile Motion

y ay vy voy t-1050 m -9.80 m/s2 0 m/s ?


y ay vy voy t-1050 m -9.80 m/s2 0 m/s ?

221 tatvy yoy 2

21 tay y

s 6.14

sm9.80

m 1050222

ya

yt

What is the final velocity vector before it hits the ground?

Does Vx change? Does Vy change? Does Vtotal change?

y ay vy voy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s

y ay vy voy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s

sm143

s 6.14sm80.90 2

tavv yoyy


tv sm143tyv

sm115txv

sm184sm143sm115 22 tv

2.51115143tan 1

Conceptual 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?

Conceptual 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?


Typical Projectile Solution Pattern:1.Calculate something in y direction. Use it to: 2. Find the time of flight.3. Find the X displacement.


The Height of a Kickoff

A placekicker kicks a football at an angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.



A placekicker kicks a football at an angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, find total time of flight.



A placekicker kicks a football at an angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, Find the range.


What did we assume that we have not yet stated???


What did we assume that we have not yet stated???Ball starts at 0 feet high.Close enough to count! That’s an engineering answer.


smvo /22

oxv

oyvDeg40

sm1440sinsm22sin ooy vv

sm1740cossm22cos oox vv


y ay vy voy t? -9.80 m/s2 0 14 m/s

3.3 Projectile Motion. Find height or ymax first.

y ay vy voy t? -9.80 m/s2 0 14 m/s

yavv yoyy 222 y

oyy

a

vvy

2

22

m 10

sm8.92

sm1402

2

y


Example 7 The Time of Flight of a Kickoff

What is the time of flight between kickoff and landing?


y ay vy voy t0 -9.80 m/s2 14 m/s ?


y ay vy voy t0 -9.80 m/s2 14 m/s ?

221 tatvy yoy

2221 sm80.9sm140 tt

t2sm80.9sm1420

s 9.2t

/ t and Clear Fractions


Example 8 The Range of a Kickoff

Calculate the range R of the projectile.

m 49s 9.2sm17

221

tvtatvx oxxox

Non-Symmetrical Projectile Motion

• Follow the general rules for projectile motion

• Break the y-direction into parts– up and down– symmetrical only back to

initial height.


• Which part of this math is symmetrical and which is not?

Note to teacherPrint out rest of these pages and give to the class. The powerpoint handout of these is already createdas a separate file to be printed out with the answer key placed on the teacher’s desk.

Conceptual Two Ways to Throw a Stone

From the top of a cliff, a person throws two stones. The stoneshave identical initial speeds, but stone 1 is thrown downwardat some angle below the horizontal and stone 2 is thrown up atthe same angle above the horizontal. Neglecting air resistance,which stone, if either, strikes the water with greater velocity?


• Find the time of flight and the range of this bowling ball we’re throwing off the roof.

2. An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. a. At what horizontal distance from the skiers should the supplies be dropped?

b. Find the magnitude of the velocity of the box as it reaches the ground.

2. An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. a. At what horizontal distance from the skiers should the supplies be dropped?

b. Find the magnitude of the velocity of the box as it reaches the ground.

vx = 80 m/sy = 250 ma= -9.80 m/s2

2yt

g 2(250)

9.8 = 7.14 s

x = vx t = (80)(7.14) = 571 m

vx = 80 m/s

vy = gt = (9.8)(7.14) = 70 m/s

2 2R x yv v v

2 280 70 = 106 m/s

3. A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff?

3. A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff?

vx = 15 m/sx = 47 mvy = 0

xv

xt

15

47 = 3.13 s

y = ½ gt2

= ½ (9.8)(3.13)2

= 48 m

4. An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:a. Its position and velocity after 8 s

4. An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:a. Its position and velocity after 8 s

vo = 100 m/s, 30t = 8 sg = - 9.8 m/s2

vox = 100 cos 30 = 86.6 m/svoy = 100 sin 30 = 50 m/s

x = vox t = 86.6(8) = 692.8 my = voy t + ½ gt2

= 50(8) + ½ (-9.8)(8)2

= 86.4 m

vx = vox = 86.6 m/svy = voy + gt = 50 + (-9.8)(8) = - 28.4 m/sV = (86.62 + -28.42)1/2

V = 91.1 m/sθ = Inv Tan (-28.4/86.6)θ = -18.2

b. The time required to reach its maximum height

c. The horizontal distance (range)

b. The time required to reach its maximum height

At the top of the path: vy = 0vy = voy + gt g

vt oy

8.9

50

= 5.1 s

c. The horizontal distance (range)

Total time T = 2t = 2(5.1) = 10.2 s

x = vox t = 86.6(10.2) = 883.7 m

5. A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.

a. At what angle with was it released?

5. A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.

a. At what angle with respect to the horizontal was it released?

vo = 15 m/st = 2 s

time to maximum height = 1 sat the top vy = 0vy = voy + gt

g

vt o sin

ov

tgsin

15

)1(8.9sin 1 = 40.8º

b. What was the maximum height achieved by the ball?

b. What was the maximum height achieved by the ball?

y = voy t +½gt2

= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2

= 4.9 m

How to do kinematics in 2-D Remember the ball shot at the monkey. Motion, force and accelerations in the X direction do not affect Y motion. And vice versa:

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