Welcome to Day 30. September 29, 2009

2. Welcome to Day 30. September 29, 2009 Objectives SWBAT Recognize projectile motion SWBAT Solve projectile motion problems for objects Catalyst 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? - 5.0m

3. Agenda • Catalyst (15min) • Review HW (10min) • Projectile motion for objects launched at an angle (15min) • Practice! (35min)

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

5. Projectile Motion All projectiles follow parabolic paths.

7. vi Vi,y Vi,x 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. HOW DO WE FIND THE X AND Y COMPONENTS OF THE Vi??

8. vi Vi,y Vi,x OUR KINEMATIC EQUATIONS!!! • vf = vi+ a∆t • vf2 = 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… Θ

9. vi Vi,y Vi,x Horizontal velocity… • Stays constant, phew. • Vx = • ∆x = vi∆t Θ

10. vi Vi,y Vi,x Vertical velocity initial: vf = vi+ a∆t • Get this in terms of y… Θ

11. vi Vi,y Vi,x Vertical velocity initial: vf2 = vi2 + 2a∆x • Get this in terms of y… Θ

12. vi Vi,y Vi,x Vertical velocity initial: ∆x = vit +½a∆t2 • Get this in terms of y… Θ

13. Angular Projectile Motion vx = vi(cosθ) = constant x = vi(cosθ)(t) vyf = vi(sinθ) + a(t) vyf2 = vi2(sinθ)2 + 2(g)(y) y = vi(sinθ)(t) + ½(g)(t)2

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

15. Projectile Motion

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

17. 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…

18. vf = vi+ a ∆t vf2 = vi2 + 2a∆x ∆x = ½(vf + vi)∆t ∆x = vit +½a∆t2

19. Projectile Motion Vertical motion of a projectile that falls from rest vyf = a(t) vyf2 = 2(a)(y) y = ½(a)(t)2

20. 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.

21. Projectile Motion Horizontal Motion of a projectile x = vx(t) vx= vx,i = constant

22. 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

23. 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

24. 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

25. 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

26. Practice • Pg 102, 1-4