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Projectile Motion

Ryan Tsuha Celena Ahn Sang Eun Lee Physics, Per 5. 2011. Projectile Motion. Vector Quantities. A quantity such as force, that has both magnitude and direction. Examples: Velocity, Acceleration. Scalar Quantities.

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Projectile Motion

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  1. Ryan Tsuha Celena Ahn Sang Eun Lee Physics, Per 5. 2011 Projectile Motion

  2. Vector Quantities A quantity such as force, that has both magnitude and direction. Examples: Velocity, Acceleration

  3. Scalar Quantities A quantity such as mass, volume, and time, which can be completely specified by it’s magnitude, and has no direction.

  4. Vector A quantity that has magnitude and direction Usually represented by an arrow whose length represents the magnitude and direction represents the direction

  5. Vector scaled 1 cm = 20 m/s This vector represents 60 m/s to the right.

  6. Resultant The vector sum of the two or more component vectors

  7. 3 Step Technique 1) Draw the two vectors with their tails touching

  8. 2) Draw a parallel projection of each vector with dashed lines to form a rectangle.

  9. 3) Draw the diagonal from the point where the two tails are touching.

  10. OR 1. Connect Draw the first vector

  11. OR 1. Connect Draw the first vector 2. Connect the tail of the second to the head of the first

  12. OR 1. Connect Draw the first vector 2. Connect the tail of the second to the head of the first 3. Draw the Diagonal

  13. Sample Question If a plane heads north at 100m/s and the wind is traveling south at 20m/s. What is the resulting velocity? 100 m/s 20 m/s

  14. Finding the magnitude of the sides Use the Pythagorean Theorem (7-m)2 + (9-m)2 = c2 130-m2 = c2 11.4-m = c 7-m 11.4-m 9-m

  15. Vectors The 3-unit and 4-unit vectors add to produce a resultant vector of 5 units, at 37.5 degrees from the horizontal

  16. Vectors The diagonal of a square is 2, times the length of one of its sides.

  17. Components of Vectors Resolution- the process of solving a vector into components

  18. Components of Vectors Any vector drawn can be resolved into vertical and horizontal components

  19. Finding the Angle from the Origin • When dealing with right triangles there are 3 trigonometric functions • SOH- Sine= Opposite/Hypotenuse • CAH- Cosine= Adjacent/Hypotenuse • TOA- Tangent= Opposite/Adjacent • Opposite Hypotenuse • AdjacentAngle Trying to find

  20. Find Angle using Tangent Tangent= Opposite/Adjacent 12 (Opposite) 5 (Adjacent) Angle θ • tan(θ) = 12/5 tan(θ) = 2.4 • θ = tan-1(2.4) How is this done on our calculators?

  21. Calculator • Make sure calculator is in DEGREES • DRG

  22. Calculator • Find Inverse Tangent of 2.4 • 2nd function • tan-1 • Enter 2.4 • Press = • 67.3 degrees 32.8 tan-1(2.4

  23. Projectile Motion Projectile- any object that moves through air or through space, acted on only by gravity

  24. Projectile Motion Projectiles near the surface of the Earth follow a curved path, due to the force of gravity.

  25. Projectile Motion Horizontal component- when no horizontal force acts on the force acts on a projectile horizontal velocity is constant

  26. Sample Question At the instant a horizontally pointed cannon is fired, a cannonball held at the canon’s side is released and drops to the ground. Which cannonball strikes the ground first, the one fired from the cannon or the one dropped?

  27. Projectiles The path traced by a projectile accelerating only in the Vertical direction while moving at a constant horizontal velocity is a parabola.

  28. Air Resistance When air resistance is small enough to neglect, usually for slow moving or very heavy projectiles, the curved paths are parabolic. Air Resistance is TOO complex for our Introduction to Physics Class

  29. Projectile Motion Vertical component- response to the force of gravity

  30. Projectile Motion IMPORTANT: The horizontal component of motion for a projectile is completely independent of the vertical component of motion.

  31. Upwardly Launched Projectiles With no gravity, the projectile will follow a straight-line path.

  32. Upwardly Launched Projectiles The horizontal range of a projectile depends on the angle of launched. The greater the launch angle, the higher the projectile will go, but…

  33. Upwardly Launched Projectiles Complimentary launch angles will travel the same horizontal distance!!!

  34. Calculations for Launch • Given the initial Velocity vi and Angleθ • A. Find Components of Velocity • 1. vv = vi sin θ • 2. vh = vi cosθ vi vv θ vh

  35. For 35o at initial velocity = 100 m/s Find Components of Velocity 1. vv = 100 m/ssin35o= 57.4 m/s 2. vh = 100 m/scos 35o= 81.9 m/s

  36. Calculations for Launch • Given the initial Velocity vi and Angleθ • A. Find Components of Velocity • 1. vv = vi sin θ • 2. vh = vi cos θ • B. Find time to get to top of path (B) • If a = (vB - vv)/t • and vB = 0 • then t = - vv/a

  37. For 35o at initial velocity = 100 m/s A. Find Components of Velocity 1. vv = 100 m/ssin35o= 57.4 m/s 2. vh = 100 m/scos 35o= 81.9 m/s B. Find time to get to top of path (B) t = - vv/a = -57.4 m/s / (-10 m/s2) t = 5.7 seconds

  38. Calculations for Launch • Given the initial Velocity vi and Angleθ • A. Find Components of Velocity B. Find time to get to top of path (B) • 1. vv = vi sin θ1. a = (vTOP - vv)/t • 2. vh = vi cos θ2. vTOP = 0 • 3. t = - vv/a • C. Height y = ½ at2 • D. Total distance traveled • x = vh(2t)

  39. For 35o at initial velocity = 100 m/s A. Find Components of Velocity 1. vv = 100 m/ssin35o= 57.4 m/s 2. vh = 100 m/scos 35o= 81.9 m/s B. Find time to get to top of path (B) t = - vv/a = -57.4 m/s / (-10 m/s2) t = 5.7 seconds y = ½ at2 = 5 m/s2(5.7 s)2 = 162 m x = vh(2t) = 81.9 m/s (11.4 s) = 934 m

  40. Calculations for Cliff • Find the time to hit the ground. • a. Measure height (y) • b. Use y = ½ atv2 or[tv= (2y/a)1/2]

  41. Calculations for Cliff • Find the time to hit the ground. • a. Measure height (y) • b. Use y = ½ atv2 or[tv= (2y/a)1/2] • Find the horizontal velocity (vh) • a. Find the time (th) to go horizontal (x) • b. vh = x/th

  42. Calculations for Cliff • Find the time to hit the ground. • a. Measure height (y) • b. Use y = ½ atv2 or[tv = (2y/a)1/2] • Find the horizontal velocity (vh) • a. Find the time (th) to go horizontal (x) • b. vh= x/th • 3. Find horizontal distance traveled from base of cliff • a. Distance = vhtv

  43. Fast-Moving Projectiles Satellite – an object that falls around the Earth or some other body rather than into it

  44. Satellites Curvature of the earth enters into our calculations If an object is moving at 8000 m/sec AND it starts falling from 5 m above the surface It will be 5 m above the ground after 1-s

  45. Satellites Throw at 8000 m/sec This is about 18,000 mph Earth circumference is 25,000 miles Takes 25000/18000 = 1.4 hours = 84 minutes Higher altitude longer

  46. Satellites Force of gravity on bowling ball is at 90o to velocity, so it doesn’t change the velocity!!!

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