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Projectile Motion

Projectile Motion. From the ground From a cliff. Kicked off a cliff. Review: Objects Kicked off a cliff have: X-dir Y-dir v= const. vi=0 m/s a= 0 m/s 2 a = -9.8 m/s Δ x= positive Δ y= negative v f = negative. Kicked off a cliff. This is a combination of basic Horizontal Motion

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Projectile Motion

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  1. Projectile Motion From the ground From a cliff

  2. Kicked off a cliff • Review: • Objects Kicked off a cliff have: X-dirY-dir v= const. vi=0 m/s a= 0 m/s2 a = -9.8 m/s Δx= positive Δy= negative vf= negative

  3. Kicked off a cliff • This is a combination of basic Horizontal Motion & Vertical Motion

  4. Projectile Motion- Level Ground • Examples: Footballs, golf balls, bullets, catapults. There is no displacement in the Δy direction between the beginning “sea level” and the ending “sea level”

  5. Projectile Motion- Level Ground • This is a combination of Vertical Motion & Horizontal Motion

  6. Projectile Motion- Level Ground • What do we know about: X-directionY-direction

  7. Projectile Motion- Level Ground • What do we know about: X-directionY-direction Δxright= + Δy= 0 v=constant vi= + a= 0m/s2vf = - a= -9.8m/s2

  8. Projectile Motion- Level Ground • Attacking the problem: Initial velocities will be given in vectors: 25 m/s @ an angle of 30˚. 30˚

  9. Projectile Motion- Level Ground • Resolve the vector into components to determine Initial Velocity in the X and Y directions. 25 m/s 30˚

  10. Projectile Motion- Level Ground • Resolve the vector into components to determine Initial Velocity in the X and Y directions. X-dirY-dir 25 m/s vy 30˚ vx

  11. Projectile Motion- Level Ground • What do we know about: X-directionY-direction Δxright= +vxΔy= 0 v=constant vi= +vy a= 0m/s2vf = -vy a= -9.8m/s2 Now you are ready to solve the problem…

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