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Welcome to Day 30. September 29, 2009

Dec 31, 2015

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Haley Bradshaw

Welcome to Day 30. September 29, 2009. Objectives SWBAT Recognize projectile motion SWBAT Solve projectile motion problems for objects Catalyst - PowerPoint PPT Presentation
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Page 1: Welcome to Day 30. September 29, 2009
Page 2: Welcome to Day 30. September 29, 2009

Welcome to Day 30. September 29, 2009

ObjectivesSWBAT Recognize projectile motionSWBAT Solve projectile motion problems for

objects

CatalystA pelican flying along a horizontal path drops a fish

from a height of 5.4 m while traveling 5.0 m/s. How far does the fish travel horizontally before it hits the water below?

- 5.0m

Page 3: Welcome to Day 30. September 29, 2009

Agenda

Catalyst (15min) Review HW (10min) Projectile motion for objects launched

at an angle (15min) Practice! (35min)

Page 4: Welcome to Day 30. September 29, 2009

HW Answers

1. .66m/s 2. 4.9m/s 3. 7.6m/s 4. 5.6m

Page 5: Welcome to Day 30. September 29, 2009

Projectile Motion

All projectiles follow parabolic paths.

Page 6: Welcome to Day 30. September 29, 2009
Page 7: Welcome to Day 30. September 29, 2009

Projectile Motion

• When a projectile is launched at an angle to the horizontal, it will have both an initial vertical component as well as a horizontal component.

vi

Vi,x

Vi,y

HOW DO WE FIND THE X AND Y COMPONENTS OF THE Vi??

Page 8: Welcome to Day 30. September 29, 2009

OUR KINEMATIC EQUATIONS!!!

vf = vi+ a∆t vf

2 = vi2 + 2a∆x

∆x = ½(vf + vi)∆t ∆x = vit +½a∆t2

But we need them to be in terms of X and Y again…

vi

Vi,x

Vi,y

Θ

Page 9: Welcome to Day 30. September 29, 2009

Horizontal velocity…

Stays constant, phew. Vx =

∆x = vi∆t

vi

Vi,x

Vi,y

Θ

Page 10: Welcome to Day 30. September 29, 2009

Vertical velocity initial: vf = vi+ a∆t

Get this in terms of y…

vi

Vi,x

Vi,y

Θ

Page 11: Welcome to Day 30. September 29, 2009

Vertical velocity initial: vf

2 = vi2 + 2a∆x

Get this in terms of y…

vi

Vi,x

Vi,y

Θ

Page 12: Welcome to Day 30. September 29, 2009

Vertical velocity initial: ∆x = vit +½a∆t2

Get this in terms of y…

vi

Vi,x

Vi,y

Θ

Page 13: Welcome to Day 30. September 29, 2009

vx = vi(cosθ) = constantx = vi(cosθ)(t)

vyf = vi(sinθ) + a(t)vyf

2 = vi2(sinθ)2 + 2(g)(y)

y = vi(sinθ)(t) + ½(g)(t)2

Angular Projectile Motion

Page 14: Welcome to Day 30. September 29, 2009

Projectile Motion

A projectile is any object propelled through space by a force. (Ex – balls, missles or arrows)

Page 15: Welcome to Day 30. September 29, 2009

Projectile Motion

Page 16: Welcome to Day 30. September 29, 2009

What do you notice about the components for the Velocity vector?

Page 17: Welcome to Day 30. September 29, 2009

So we can still use our Kinematic equations!

We just need to adjust them so that we solve for only X components and then only for Y components

Today we will only talk about objects that have no initial vertical velocity

Vertical motion of a projectile that falls from rest derivation…

Page 18: Welcome to Day 30. September 29, 2009

vf = vi+ a ∆t

vf2 = vi

2 + 2a∆x∆x = ½(vf + vi)∆t

∆x = vit +½a∆t2

Page 19: Welcome to Day 30. September 29, 2009

Projectile Motion

Vertical motion of a projectile that falls from rest

vyf = a(t)vyf

2 = 2(a)(y)y = ½(a)(t)2

Page 20: Welcome to Day 30. September 29, 2009

Horizontal motion of an object that falls from rest…

This one is easy… If we neglect air resistance, our object

will remain constant throughout So, the initial horizontal velocity is the

horizontal velocity throughout the entire flight.

Page 21: Welcome to Day 30. September 29, 2009

Projectile Motion

Horizontal Motion of a projectile

x = vx(t)vx= vx,i = constant

Page 22: Welcome to Day 30. September 29, 2009

The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a little rock horizontally off the bridge. The rock hits the water such that the magnitude of its horizontal displacement is 45.0m. Find the speed at which the rock was kicked.

5.56m/s

Page 23: Welcome to Day 30. September 29, 2009

A plane flies horizontally at 50.0 m/s. If the plane is 54.0 m above the ground, how far will the package travel horizontally before it hits the ground?

166 m

Page 24: Welcome to Day 30. September 29, 2009

A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table at a speed of 5.0 m/s. Where does the cat strike the floor?

2.3 m from the table

Page 25: Welcome to Day 30. September 29, 2009

A pelican flying along a horizontal path drops a fish from a height of 5.4 m while traveling 5.0 m/s. How far does the fish travel horizontally before it hits the water below?

5.0 m

Page 26: Welcome to Day 30. September 29, 2009

Practice

Pg 102, 1-4