2-Dimensional Motion2-Dimensional MotionHorizontalHorizontal VerticalVertical

The only thing to affect the ball is friction, and we don’t look at that!

Velocity is always constant!

What affects the ball?Gravity! (No friction)

g = -9.8m/s2

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CombinCombine the e the twotwo

Horizontal and Vertical components are independent!

(They don’t affect each other. It doesn’t matter how fast the object is going in the x direction, gravity still acts evenly on the object.) [Quarters]

So … all these problems are at 2 problems: one problem in the y direction, one problem in the x direction.

time is what combines the two.

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Steps for all these problemsSteps for all these problems

½.½. Picture & Givens Picture & Givens 1.1. Break the velocity into x and y componentsBreak the velocity into x and y components

vvixix=v=vi i cos cos θθ, v, viyiy=v=vi i sin sin θθ2.2. Use the x or y displacement to find time Use the x or y displacement to find time

(x = v(x = vi i t + ½att + ½at22))3.3. Use the time to find the other displacement: Use the time to find the other displacement:

x or y. (x = vx or y. (x = viit + ½att + ½at22))

4.4. Find the maximum height. (vFind the maximum height. (vff22 = v = vii

22 +2ax) +2ax)5.5. Find the final velocity.Find the final velocity.

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Sample Problem #1Sample Problem #1

A plane flying horizontally (0°) at 91 m/s (204 A plane flying horizontally (0°) at 91 m/s (204 mph) drops a bag of food supplies from an mph) drops a bag of food supplies from an altitude of 812 m (0.5 mile). Find:altitude of 812 m (0.5 mile). Find:

a. Picture & Givensa. Picture & Givens

b. Initial x & y velocitiesb. Initial x & y velocities

c. Time required for the food supplies to fall.c. Time required for the food supplies to fall.

d. Range of the projectile (x distance)d. Range of the projectile (x distance)

e. Final Velocity of the food supplies just e. Final Velocity of the food supplies just before hitting the ground.before hitting the ground.

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Givens:Givens:

vvi i = 91 m/s= 91 m/s

y = -812 my = -812 m

aayy = -9.8 m/s = -9.8 m/s22

1. Break velocity into components.

vix = 91 cos 0 = 91

viy = 91 sin 0 = 02. Use disp. to find time.

x = vit + ½at2

-812=0 + ½ (-9.8)t2

t = 12.9 sec

3. Use time to find disp.x = vix

t

x = 91 (12.9) = 1174 m4. Maximum HeightEasy in this problem! y = 812 m

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5. Find final velocity

vix = vfx

remember x velocity stays constant

vfy = viy

+ at remember y velocity is affected

by gravity

vfx

vfy vf

vfx= 91m/s

vfy = viy

+ at

vfy = 0 + -9.8

(12.9)vfy

= -126vf = √(vfy2 +vfx

2)

vf = 155 m/s

1

1

tan

91tan

12635.8

35.8 270 306

fx

fy

v

v

vf = 155 m/s at 306°

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Sample Problem #2Sample Problem #2

An arrow is shot from a bow with an initial An arrow is shot from a bow with an initial velocity of 35 m/s and an initial angle of 30velocity of 35 m/s and an initial angle of 30. . If the arrow lands in a target at the same If the arrow lands in a target at the same height it was shot, find:height it was shot, find:

a. How long is the arrow in the air? How long is the arrow in the air? b.b. How far would a target be placed in order How far would a target be placed in order

for the arrow to hit the bulls eye?for the arrow to hit the bulls eye?c.c. the maximum height the arrow travels?the maximum height the arrow travels?d.d. the arrow’s final velocity?the arrow’s final velocity?

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Givens:Givens:

vvi i = 35 m/s= 35 m/s

y = 0 my = 0 m

aayy = -9.8 m/s = -9.8 m/s22

1. Break velocity into components.

vix = 35 cos 30 = 26.0 m/s

viy = 35 sin 30 = 17.5 m/s2. Use disp. to find time.

x = vit + ½at2

0 = (17.5)t+ ½ (-9.8)t2

t = 1.89 sec

3. Use time to find disp.x = vix

t

x = 26.0 (1.89) = 49.1 m

4. Maximum Heightvf

2 = vi2 +2ax

0 = 17.52 + 2(-9.8) y

y = 15.6 m

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Sample Problem #3Sample Problem #3

A disgruntled physics student throw A disgruntled physics student throw their book off a 10 m high bridge. If their book off a 10 m high bridge. If they throw it at 5.4 m/s at 28°, then they throw it at 5.4 m/s at 28°, then find:find:a. va. vixix & v & viyiy

b. time in the airb. time in the air

c. range (x-disp)c. range (x-disp)

d. maximum heightd. maximum height

e. final velocitye. final velocity

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a. va. vixix = v = vii cos cos θθ = 5.4 cos 28 = 4.77 m/s = 5.4 cos 28 = 4.77 m/s

vviyiy = v = vii sin sin θθ = 5.4 sin 28 = 2.54 m/s = 5.4 sin 28 = 2.54 m/sb. y = vb. y = viit + ½att + ½at22

-10 = 2.54 t + ½(-9.8)t-10 = 2.54 t + ½(-9.8)t22

0 = -4.9t2 + 2.54t +100 = -4.9t2 + 2.54t +10

Quadratic Equation

2 4

2

b b ac

a

22.54 2.54 4( 4.9)(10)

2( 4.9)

2.54 14.2

2( 4.9)

1.19 or 1.71t

c. x = vt

x = 4.77(1.71)

x = 8.16 m(Go with the + time or the greater time!)

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d. maximum heightd. maximum height

vvff22 = v = vii

22 + 2ax + 2ax

0 = 2.540 = 2.5422 +2(- +2(-9.8)y9.8)y

y = 0.329 my = 0.329 m

e. final velocitye. final velocity

vvfxfx = v = vixix = 4.77 m/s = 4.77 m/s

vvfyfy = v = viyiy + at + at

= 2.54 + -9.8 (1.71)= 2.54 + -9.8 (1.71)

vvfyfy = -14.2 m/s = -14.2 m/s

14.2

4.772 24.77 14.2 15.0 m/s

4.77=tan-1 18.6 270 288

14.2

vf