Kalkulus II (17 - 18)

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Kalkulus II (17 - 18)

  1. 1. Kalkulus II<br />Teguh Budi P, M.Si <br />Sesion #17-18<br />JurusanFisika<br />FakultasMatematikadanIlmuPengetahuanAlam<br />
  2. 2. Outline<br />Projectile Motion<br />1/10/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />2<br />
  3. 3. Vectors and Motion in Spacepart (1)<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />3<br />1/10/2011<br />
  4. 4. One early use of calculus was to study projectile motion.<br />In this section we assume ideal projectile motion:<br />Constant force of gravity in a downward direction<br />Flat surface<br />No air resistance (usually)<br />1/10/2011<br />4<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  5. 5. We assume that the projectile is launched from the origin at time t=0 with initial velocity vo.<br />The initial position is:<br />1/10/2011<br />5<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  6. 6. Newton’s second law of motion:<br />Vertical acceleration<br />1/10/2011<br />6<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  7. 7. Newton’s second law of motion:<br />The force of gravity is:<br />Force is in the downward direction<br />1/10/2011<br />7<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  8. 8. Newton’s second law of motion:<br />The force of gravity is:<br />1/10/2011<br />8<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  9. 9. Newton’s second law of motion:<br />The force of gravity is:<br />1/10/2011<br />9<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  10. 10. Initial conditions:<br />1/10/2011<br />10<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  11. 11. Vector equation for ideal projectile motion:<br />1/10/2011<br />11<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  12. 12. Vector equation for ideal projectile motion:<br />Parametric equations for ideal projectile motion:<br />1/10/2011<br />12<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  13. 13. Example 1:<br />A projectile is fired at 60o and 500 m/sec.<br />Where will it be 10 seconds later?<br />The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers.<br />Note: The speed of sound is 331.29 meters/sec<br />Or 741.1 miles/hr at sea level.<br />1/10/2011<br />13<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  14. 14. The maximum height of a projectile occurs when the vertical velocity equals zero.<br />time at maximum height<br />1/10/2011<br />14<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  15. 15. The maximum height of a projectile occurs when the vertical velocity equals zero.<br />We can substitute this expression into the formula for height to get the maximum height.<br />1/10/2011<br />15<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  16. 16. 1/10/2011<br />16<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  17. 17. maximum<br />height<br />1/10/2011<br />17<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  18. 18. When the height is zero:<br />time at launch:<br />1/10/2011<br />18<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  19. 19. When the height is zero:<br />time at launch:<br />time at impact<br />(flight time)<br />1/10/2011<br />19<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  20. 20. If we take the expression for flight time and substitute it into the equation for x, we can find the range.<br />1/10/2011<br />20<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  21. 21. If we take the expression for flight time and substitute it into the equation for x, we can find the range.<br />Range<br />1/10/2011<br />21<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  22. 22. The range is maximum when<br />is maximum.<br />Range is maximum when the launch angle is 45o.<br />Range<br />1/10/2011<br />22<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  23. 23. If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.<br />1/10/2011<br />23<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  24. 24. If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.<br />This simplifies to:<br />which is the equation of a parabola.<br />1/10/2011<br />24<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  25. 25. If we start somewhere besides the origin, the equations become:<br />1/10/2011<br />25<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  26. 26. Example 4:<br />A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20o from the horizontal. A gust of wind adds a component of -8.8 ft/sec in the horizontal direction to the initial velocity.<br />The parametric equations become:<br />1/10/2011<br />26<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  27. 27. In real life, there are other forces on the object. The most obvious is air resistance.<br />If the drag due to air resistance is proportional to the velocity:<br />(Drag is in the opposite direction as velocity.)<br />Equations for the motion of a projectile with linear drag force are given on page 546.<br />You are not responsible for memorizing these formulas.<br />1/10/2011<br />27<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
  28. 28. Thank You<br />1/10/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />28<br />

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