Welcome to Day 30. September 29, 2009
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Transcript of Welcome to Day 30. September 29, 2009
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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
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Agenda
Catalyst (15min) Review HW (10min) Projectile motion for objects launched
at an angle (15min) Practice! (35min)
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HW Answers
1. .66m/s 2. 4.9m/s 3. 7.6m/s 4. 5.6m
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Projectile Motion
All projectiles follow parabolic paths.
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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??
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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
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Horizontal velocity…
Stays constant, phew. Vx =
∆x = vi∆t
vi
Vi,x
Vi,y
Θ
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Vertical velocity initial: vf = vi+ a∆t
Get this in terms of y…
vi
Vi,x
Vi,y
Θ
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Vertical velocity initial: vf
2 = vi2 + 2a∆x
Get this in terms of y…
vi
Vi,x
Vi,y
Θ
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Vertical velocity initial: ∆x = vit +½a∆t2
Get this in terms of y…
vi
Vi,x
Vi,y
Θ
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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
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Projectile Motion
A projectile is any object propelled through space by a force. (Ex – balls, missles or arrows)
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Projectile Motion
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What do you notice about the components for the Velocity vector?
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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…
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vf = vi+ a ∆t
vf2 = vi
2 + 2a∆x∆x = ½(vf + vi)∆t
∆x = vit +½a∆t2
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Projectile Motion
Vertical motion of a projectile that falls from rest
vyf = a(t)vyf
2 = 2(a)(y)y = ½(a)(t)2
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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.
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Projectile Motion
Horizontal Motion of a projectile
x = vx(t)vx= vx,i = constant
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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
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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
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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
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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
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Practice
Pg 102, 1-4