Quadratic Cliff Jumping Deann Anguiano Laura Moore-Mueller Russ Ballard Lake Chelan Conference 2008.
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Quadratic Cliff Quadratic Cliff Jumping Jumping Deann Anguiano Laura Moore-Mueller Russ Ballard Lake Chelan Conference 2008
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3 Motion along a curved path What is happening The scene is messy What can be assumed What are the concepts
Transcript of Quadratic Cliff Jumping Deann Anguiano Laura Moore-Mueller Russ Ballard Lake Chelan Conference 2008.
Quadratic Cliff JumpingQuadratic Cliff Jumping
Deann AnguianoLaura Moore-Mueller
Russ Ballard Lake Chelan Conference 2008
May 2, 2008
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May 2, 2008
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Motion along a curved path
•What is happening
•The scene is messy
•What can be assumed
•What are the concepts
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Why use projectile motion?
•Position vs. position graph
•Apply regression curve
•First few points do not work
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Which points? •Depends on what’s occurring
•Symmetry
•Vertex
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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
May 2, 2008
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Challenge cup
•Team Crimson
– Given a height you pick position
•Team Silver
– Given a position you pick the height
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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
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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?
May 2, 2008
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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.
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What is the motion?•In the horizontal?
•In the vertical?
•Motion is parametric
•Look at the vertical motion
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Horizontal motion
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Vertical motion
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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
May 2, 2008
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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)
I second
Velocity = -9.81 m/sVelocity(m/s)
time (s)
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Y = At^2 + Bt +CA:-4.844B:6.513C:-1.168
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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
May 2, 2008
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Vernier LoggerPRo•Hand out Vernier LoggerPRo
profile
May 2, 2008
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Video Analysis•Demo video analysis of the
clip if time
All info is located at russballard.com/workshop/chelan conference 2008
Questions?
