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Understanding Projectile Motion: From Ground to Cliff

Explore the principles of projectile motion when objects are launched from the ground or kicked off a cliff. This review covers the key concepts in both horizontal and vertical motion, with examples like footballs, golf balls, bullets, and catapults. Learn how to resolve velocity vectors and tackle problems related to projectile motion effectively.

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Understanding Projectile Motion: From Ground to Cliff

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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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