09-projectile motion

09-projectile motion - PHYS 211 2/02/09 2D and 3D Motion...

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Unformatted text preview: PHYS 211 2/02/09 2D and 3D Motion Describe: Displacement, Velocity, and Acceleration Projectile Motion Example Particle has velocity Gets an acceleration a=3.0 m/s 2 at an angle of 130 Find the magnitude and direction of the velocity vector 5 seconds later. v = 2.0 i 4.0 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 + a t Example Continued a v v f a = v f- v v = v + a t v x = v x + a x t v y = v y + a y t Vector equation represents two equations a x = a cos( ) = 3.0( m / s 2 )cos(130 ) = 1.93 m / s 2 a y = a sin( ) = 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 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 i + 7.5 j ( m / s ) = tan 1 7.5 11.6 = 33 | v f | = 11.6 2 + 7.5 2 = 13.8 ( m / s ) = 147 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?...
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This note was uploaded on 10/03/2011 for the course PHYS 211 taught by Professor Pettie during the Spring '08 term at South Carolina.

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09-projectile motion - PHYS 211 2/02/09 2D and 3D Motion...

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