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Chapter 3 – 2D Motion

Chapter 3 – 2D Motion. Definitions Projectile motion. Find the horizontal and vertical components of the d = 140 m displacement of a superhero who flies from the top of a tall building following the path shown in the Figure where = 30.0°. Superperson. 2 Dimensional Motion.

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Chapter 3 – 2D Motion

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  1. Chapter 3 – 2D Motion • Definitions • Projectile motion

  2. Find the horizontal and vertical components of the d = 140 m displacement of a superhero who flies from the top of a tall building following the path shown in the Figure where = 30.0°. Superperson

  3. 2 Dimensional Motion • We will consider motion the the x-y plane. • Positions now have (x,y) coordinates so we need to use vectors. • There are two types of problems we need to consider • Throw or drop an object at an angle to the horizontal • Make something go around in a circle

  4. Definitions Position vector: extends from the origin of a coordinate system to the particle. Displacement vector: represents a particle’s position change during a certain time interval. Average velocity:

  5. Instantaneous velocity: • The direction of the instantaneous velocity of a • particle is always tangent to the particle’s path at • the particle’s position Average acceleration: Instantaneous acceleration:

  6. Instantaneous velocity: • The direction of the instantaneous velocity of a • particle is always tangent to the particle’s path at • the particle’s position Average acceleration: Instantaneous acceleration:

  7. Projectile Motion What is its velocity here? It’s acceleration? How long did it take to get here? here? height (h) RANGE (R)

  8. Motion of a particle launched with initial velocity, v0 and free fall acceleration g. II. Projectile motion The horizontal and vertical motions are independent from each other. - Horizontal motion:ax=0  vx=v0x= cte Range (R):horizontal distance traveled by a projectile before returning to launch height. - Vertical motion:ay= -g

  9. - Trajectory: projectile’s path. - Horizontal range: R = x-x0; y-y0=0. (Maximum for a launch angle of 45º ) Overall assumption: the air through which the projectile moves has no effect on its motion  friction neglected.

  10. In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement. y v0 θ=45º x x=R=R’?

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