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09-projectile motion

# 09-projectile motion - PHYS 211 2D and 3D Motion Describe...

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PHYS 211 2/02/09 2D and 3D Motion Describe: Displacement, Velocity, and Acceleration Projectile Motion

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Example Particle has velocity Gets an acceleration a=3.0 m/s 2 at an angle of 130º Find the magnitude and direction of the velocity vector 5 seconds later. v = 2.0 ˆ i 4.0 ˆ j ( m / s ) Steps: Sketch vectors on graph Resolve a into components a x and a y Compute new velocity in each dimension independently Use components of new v to get magnitude and direction v = v 0 + a t
Example Continued a v v f a = v f - v v = v 0 + a t v x = v 0 x + a x t v y = v 0 y + a y t Vector equation represents two equations a x = a cos( θ ) = 3.0( m / s 2 )cos(130 ° ) = 1.93 m / s 2 a y = a sin( θ ) = 3.0( m / s 2 )sin(130 ° ) = 2.3 m / s 2 a x a y v x v y v v f v fx v fy θ v x = 2 ( m / s ) + ( 1.93 m / s )(5 s ) = 11.6 m / s v y = 4 ( m / s ) + (2.3 m / s )(5 s ) = 7.5 m / s v f = 11.6 ˆ i + 7.5 ˆ j ( m / s ) θ = tan 1 7.5 11.6 = 33 ° | v f | = 11.6 2 + 7.5 2 = 13.8 ( m / s ) θ = 147 °

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Velocity and Acceleration A ball is rolling down one hill and up another as shown. Points A and B are at the same height. How do the velocity and acceleration change as the ball rolls from point A to point B? a) The velocity and acceleration are the same at both points. b) The velocity and the magnitude of the acceleration are the same at both points, but the direction of the acceleration is opposite at B to the direction it had at A.
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09-projectile motion - PHYS 211 2D and 3D Motion Describe...

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