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Quadratic Cliff Jumping. Deann Anguiano Laura Moore-Mueller Russ Ballard Lake Chelan Conference 2008. Motion along a curved path . What is happening The scene is messy What can be assumed What are the concepts. Why use projectile motion?. Position vs. position graph

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1. Quadratic Cliff Jumping Deann Anguiano Laura Moore-Mueller Russ Ballard Lake Chelan Conference 2008

2. Motion along a curved path • What is happening • The scene is messy • What can be assumed • What are the concepts

3. Why use projectile motion? • Position vs. position graph • Apply regression curve • First few points do not work

4. Which points? • Depends on what’s occurring • Symmetry • Vertex

5. Hot wheels • Instructions • You have a sheet for the activity • Data gathering • Gather as much data as possible • Three to four runs is a minimum

6. Challenge cup • Team Crimson • Given a height you pick position • Team Silver • Given a position you pick the height

7. Tossing the ball • Watch the ball being tossed into the air • Plot a d vs. t graph of the motion on your ball toss graph

8. Tossing the ball • Exchange with a neighbor • Answer questions based on graph in front of you • How long was the ball in the air? • Is the vertical distance as the ball travels upward the same or different as the vertical distance traveled downward? • Is the time from the launch to the vertex the same as the vertex to the ground?

9. Why use distance vs time? • Return the graph to the owner and owners now reflect • What kind of motion did I graph? • If it was incorrect why would the graph mislead a student? • If necessary redraw the graph on the opposite side of the paper.

10. What is the motion? • In the horizontal? • In the vertical? • Motion is parametric • Look at the vertical motion

11. Horizontal motion

12. Vertical motion

13. a(t) = -9.81 m/sec2 v(t) = area in graph in the rectangle Therefore v(t) = -9.81m/s2 * t(s) Or the ball is traveling at -9.81*t m/s after t seconds

14. I second time (s) Velocity = -9.81 m/s Velocity(m/s) Initial velocity is zero at the top After one second it = -9.81 m/s Slope is constant so the area under the curve is a triangle =½ base * height =½ t(s) * (-9.81*t m/s) = -4.9 t2 (m)

15. Y = At^2 + Bt +C A:-4.844 B:6.513 C:-1.168

16. T3 clip • Calculate if the TX will make to the hearse roof. • Distance and speed of hearse 20 m/s 26 m travel • Distance vertical that TX will fall 8.3 m • Distance at speed horizontal she is traveling 7.9 m/s 10.3 m

17. Vernier LoggerPRo • Hand out Vernier LoggerPRo profile

18. Video Analysis • Demo video analysis of the clip if time

19. All info is located at russballard.com/workshop/chelan conference 2008

20. Questions?

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