unit 3: projectile & 2d motion

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Unit 3: Projectile & 2D Motion http://www.youtube .com/watch?v=sFcsM DzW1Yg

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Unit 3: Projectile & 2D Motion. http://www.youtube.com/watch?v=sFcsMDzW1Yg. Projectiles. Airborne objects that move as a result of their own inertia and gravity. They exhibit two dimensional motion. Horizontal and vertical motion - PowerPoint PPT Presentation

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Page 1: Unit 3:  Projectile & 2D Motion

Unit 3: Projectile & 2D

Motionhttp://www.youtube.com/watch?v=sFcsMDzW1Yg

Page 2: Unit 3:  Projectile & 2D Motion

Projectiles Airborne objects that move as a result

of their own inertia and gravity. They exhibit two dimensional motion.

Horizontal and vertical motion Horizontal and vertical motion are

independent of one another. The resultant motion is a combination of

horizontal and vertical motion.

Page 3: Unit 3:  Projectile & 2D Motion

2 Dimensional Motion Projectiles

Gravity affects outcome

Riverboats & Airplanes Gravity does not affect outcome

Page 4: Unit 3:  Projectile & 2D Motion
Page 5: Unit 3:  Projectile & 2D Motion

If a cannonball is projected horizontally from the top of a cliff. In the absence of gravity, the cannonball would continue its horizontal motion at a constant velocity. Horizontal velocity is constant (ax = 0)

If dropped from rest in the presence of gravity, the cannonball would accelerate downward. Vertical velocity gains speed (ay = -9.8 m/s2)

Page 6: Unit 3:  Projectile & 2D Motion

Monkey In the Absence of Gravity

Page 7: Unit 3:  Projectile & 2D Motion

Why Does the Hunter Miss the Monkey?

Page 8: Unit 3:  Projectile & 2D Motion

The Hunter Now Aims

Page 9: Unit 3:  Projectile & 2D Motion

Strategies for Applying Kinematic Equations in 2 Dimensions

Make a drawing of the situation.

Decide which directions to be called positive and negative. Do this for both the x and y directions.

Remember that the time variable is the same for both x and y.

Make note of all the assumed information.

Starts at rest.

Stops

When motion is divided into two segments, remember the final velocity for one segment becomes the initial velocity for the next segment.

Make notice of the situation…do you multiply or divide the time by 2… or leave it?

Page 10: Unit 3:  Projectile & 2D Motion

Anytime the object starts at a given height

Do not multiply time by 2….since starts at max height already

Page 11: Unit 3:  Projectile & 2D Motion

Launch velocity

Launch angle

Ending velocity

Anytime an object starts at an angle from the ground! Vy at the highest peak is always 0 Remember when use max height and Vy=0….it’s only

half the time the object stays in the air So…to find total distance (dx)…multiply time by 2

Page 12: Unit 3:  Projectile & 2D Motion

Steps to working a problem Draw the picture Label with variables and numbers Write the givens Write what you are looking for Write equation Plug and chug Answer with number and unit!

Page 13: Unit 3:  Projectile & 2D Motion

EX 1: A football is kicked at an angle of 45º to the ground with a speed of 25 m/s. Calculate …

A. The initial horizontal velocity B. The initial vertical velocity C. Maximum height reached

Page 14: Unit 3:  Projectile & 2D Motion

EX 2: A launched projectile has a horizontal displacement of 53 meters and total time 7 seconds. Calculate … … note – think of various ways this question could be formed on a test (cannon fires, football thrown, baseball hit, golf ball hit, soccer ball kicked, etc.) A. Maximum height of projectile B. Launch Velocity C. Launch angle

Page 15: Unit 3:  Projectile & 2D Motion

A launched projectile has a horizontal displacement of 53 meters and total time 7 seconds. Calculate …

A. Maximum height of projectile.Answer … -Begin with equation-Use the fact that at top of trajectory Vyi is

zero. -Use second half of trajectory.-Plugging in values:

2y at

2

1tvd i

mss

md y 025.60)5.3()8.9(2

1 22

Page 16: Unit 3:  Projectile & 2D Motion

A launched projectile has a horizontal displacement of 53 meters and total time 7 seconds. Calculate … B. Launch Velocity.Answer … Need X and Y velocities (Vx and Vy). Vx is found using:

Vy is found using half of trajectory and:Knowing that Vy is zero at half the time …

So …

sms

m

t

dv xx 57.7

7

53

sms

smatvv iyy 3.34)5.3)(8.9(0 2

smvvVo xy 125.35)()( 22

Page 17: Unit 3:  Projectile & 2D Motion

A launched projectile has a horizontal displacement of 53 meters and total time 7 seconds. Calculate …

C. Launch Angle.

smVy 3.34

θ

smVx 57.75.77)

57.7

3.34(tan 1

Page 18: Unit 3:  Projectile & 2D Motion

Ex 3: An object is launched horizontally from a height of 350 meters and lands 900 meters horizontally. Calculate …

Velocity of object.

s

smm

g

dt y 45.8

)8.9(

)350(22

2

sms

m

t

dv xx 5.106

45.8

900

Page 19: Unit 3:  Projectile & 2D Motion

Ex 4 An object is travelling at 37◦ from vertical at velocity 8 m/s. How long will it take to go 38 m horizontally? NOT A PROJECTILE!!!! Need X component velocity.Note that (similar to projectiles) time is shared between

two motions. Each component is independent of the other.

smVo 8

θ=37

sm

smvx 39.6)37cos()8(

s

smm

v

dt

x

x 95.539.6

38

xv

yv

Page 20: Unit 3:  Projectile & 2D Motion

Review Questions1. In the absence of gravitational force

and friction, describe the motion of an object.

2. How do you describe horizontal motion?

3. Is vertical motion influenced by gravitational force?

4. How far is an object being thrown horizontally given an initial speed of 20 m/s from a 45 m tower?