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Projectile Motion

Projectile Motion. Independence of vectors. When a vector is moving in two dimensions the vertical and horizontal components are independent from each other. They have no affect on each other. “Cliff Problems”. V h only. “Cliff Problems”. V h only. “Cliff Problems”. V h only. V h.

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Projectile Motion

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  1. Projectile Motion

  2. Independence of vectors • When a vector is moving in two dimensions the vertical and horizontal components are independent from each other. • They have no affect on each other

  3. “Cliff Problems” Vh only

  4. “Cliff Problems” Vh only

  5. “Cliff Problems” Vh only Vh Vv

  6. “Cliff Problems • How long object is in the air is only dependent on the height of the “cliff” • All objects fall at the same rate, so it doesn’t matter if the horizontal velocity is 100 m/s or 0 m/s, the time in the air will be the same. • The object’s horizontal velocity is constant, it doesn’t accelerate horizontally, only vertically

  7. “Cliff Problems” Vh only h range

  8. “Cliff Problems” To find the time in the air use the following equation: h = gt2 2 Vh only h range

  9. “Cliff Problems” Vh only To find the range, use the time in the air and the following equation: range = (Vh)(t) h range

  10. “Cliff Problems” Vh only A cliff is 44 m high. If a car is traveling 15 m/s, how far from the base of the cliff will it land. (Find the range) h range

  11. “Cliff Problems” Vh only A cliff is 44 m high. If a car is traveling 15 m/s, how far from the base of the cliff will it land. (Find the range) • Find time in air. • h = gt2 • 2 • 44 m = (9.8m/s2)(t2) t = 2.9 sec • 2 h range

  12. “Cliff Problems” Vh only A cliff is 44 m high. If a car is traveling 15 m/s, how far from the base of the cliff will it land. (Find the range) • Calculate range: • range = Vht • = (15 m/s)(2.9 sec) • = 43.5 m h range

  13. “Shooting at an Angle” • This type of problem deals with the launching of an object at an angle. • The object has both vertical and horizontal velocities. • These are independent of each other.

  14. “Shooting at an Angle” Vv Vh

  15. “Shooting at an Angle” • The horizontal velocity remains constant • The vertical velocity slows down due to gravity and then accelerates as it comes back down • They are independent of each other

  16. “Shooting at an Angle” velocity Vv θ Vh A gun is fired at a 66 ° with a speed of 4.47 m/s. Find how long the bullet is in the air, how high, and how far it travels.

  17. “Shooting at an Angle” velocity Vv θ Vh First, find the vertical and horizontal components Vertical Sin 66 ° = Vv Horizontal Cos 66 ° = Vh 4.47 m/s 4.47 m/s Vv = 4.08 m/s Vh = 1.82 m/s

  18. “Shooting at an Angle” velocity 4.08 m/s θ 1.82 m/s Next, you need to fine the time in the air. tup = Vv = 4.08 m/s = 0.416 sec g 9.8 m/s2 ttotal = .416 x 2 = 0 .833 sec.

  19. “Shooting at an Angle” velocity 4.08 m/s θ 1.82 m/s Once the time is known, you can calculate the range and the height Range = Vh x ttotal = (1.82 m/sec.)(0.833 sec) = 1.52 m 2

  20. “Shooting at an Angle” velocity 4.08 m/s θ 1.82 m/s Once the time is known, you can calculate the range and the height Height = Vvtup – ½gtup2 = (4.08 m/s)(.416 sec) – (9.8 m/s2)(0.416 sec)2 2 = .849 m

  21. Simulations http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/mzng.html

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