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PROJECTILE MOTION Senior High School Physics

PROJECTILE MOTION Senior High School Physics. Mr. T. P. Eagan 2012. Part 1. Part 2. Free powerpoints at http://www.worldofteaching.com. Introduction. Projectile Motion: Motion through the air without a propulsion Examples:.

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PROJECTILE MOTION Senior High School Physics

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  1. PROJECTILE MOTIONSenior High School Physics Mr. T. P. Eagan 2012 Part 1. Part 2. Free powerpoints at http://www.worldofteaching.com

  2. Introduction • Projectile Motion: Motion through the air without a propulsion • Examples:

  3. Part 1.Motion of Objects Projected HorizontallyThis presentation is dedicated to all of David’s pencil’s that have gone misplaced for the last four years (in class)!!! In no way would I ever dedicate anything to Beeton!!!!

  4. y v0 x

  5. y x

  6. y x

  7. y x

  8. y x

  9. y • Motion is accelerated • Acceleration is constant, and downward • a = g = -9.81m/s2 • The horizontal (x) component of velocity is constant • The horizontal and vertical motions are independent of each other, but they have a common time g = -9.81m/s2 x

  10. ANALYSIS OF MOTION • ASSUMPTIONS: • x-direction (horizontal): uniform motion • y-direction (vertical): accelerated motion • no air resistance • QUESTIONS: • What is the trajectory? • What is the total time of the motion? • What is the horizontal range? • What is the final velocity?

  11. y v0 g h x 0 Frame of reference: Equations of motion:

  12. y h v02 > v01 v01 x Trajectory x = v0 t y = h + ½ g t2 Parabola, open down Eliminate time, t t = x/v0 y = h + ½ g (x/v0)2 y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h

  13. y h Δt = √ 2h/(9.81ms-2) Δt = √ 2h/(-g) x Total Time, Δt Δt = tf - ti y = h + ½ g t2 final y = 0 0 = h + ½ g (Δt)2 ti =0 Solve for Δt: tf =Δt Total time of motion depends only on the initial height, h

  14. y h Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δx x Horizontal Range, Δx x = v0 t final y = 0, time is the total time Δt Δx = v0 Δt Horizontal range depends on the initial height, h, and the initial velocity, v0

  15. v v = √vx2 + vy2 = √v02+g2t2 tg Θ = vy/ vx = g t / v0 VELOCITY vx = v0 Θ vy = g t

  16. Δt = √ 2h/(-g) tg Θ = g Δt / v0 = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 v v = √vx2 + vy2 v = √v02+g2(2h /(-g)) v = √ v02+ 2h(-g) FINAL VELOCITY vx = v0 Θ vy = g t Θ is negative (below the horizontal line)

  17. HORIZONTAL THROW - Summary h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2

  18. Part 2.Motion of objects projected at an angle

  19. y Initial velocity: vi = vi [Θ] Velocity components: x- direction : vix = vi cos Θ y- direction : viy = vi sin Θ viy vi θ x vix Initial position: x = 0, y = 0

  20. y a = g = - 9.81m/s2 • Motion is accelerated • Acceleration is constant, and downward • a = g = -9.81m/s2 • The horizontal (x) component of velocity is constant • The horizontal and vertical motions are independent of each other, but they have a common time x

  21. ANALYSIS OF MOTION: • ASSUMPTIONS • x-direction (horizontal): uniform motion • y-direction (vertical): accelerated motion • no air resistance • QUESTIONS • What is the trajectory? • What is the total time of the motion? • What is the horizontal range? • What is the maximum height? • What is the final velocity?

  22. Equations of motion:

  23. Equations of motion:

  24. Trajectory x = vi t cos Θ y = vi tsin Θ + ½ g t2 Parabola, open down y Eliminate time, t t = x/(vi cos Θ) y = bx + ax2 x

  25. 0 =vi sin Θ + ½ g Δt Δt = 2 vi sinΘ (-g) Total Time, Δt y = vi tsin Θ + ½ g t2 final height y = 0, after time interval Δt 0 =vi Δtsin Θ + ½ g (Δt)2 Solve for Δt: x t = 0 Δt

  26. 2 vi sinΘ Δt = (-g) Δx = 2vi 2 sin Θ cos Θ vi 2 sin (2Θ) Δx = (-g) (-g) Horizontal Range, Δx x = vi t cos Θ y final y = 0, time is the total time Δt Δx = vi Δt cos Θ x 0 sin (2Θ) = 2 sin Θ cos Θ Δx

  27. vi 2 sin (2Θ) Δx = (-g) Horizontal Range, Δx • CONCLUSIONS: • Horizontal range is greatest for the throw angle of 450 • Horizontal ranges are the same for angles Θ and (900 – Θ)

  28. Trajectory and horizontal range vi = 25 m/s

  29. Velocity • Final speed = initial speed (conservation of energy) • Impact angle = - launch angle (symmetry of parabola)

  30. tup= hmax =vi t upsin Θ + ½ g tup2 hmax =vi2sin2Θ/(-g) + ½ g(vi2sin2Θ)/g2 hmax = vi sin Θ (-g) vi2sin2Θ 2(-g) Maximum Height vy = vi sin Θ + g t y = vi tsin Θ + ½ g t2 At maximum height vy = 0 0 = vi sin Θ + g tup tup= Δt/2

  31. vi 2 sin (2Θ) (-g) Projectile Motion – Final Equations (0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 2 vi sinΘ (-g) vi2sin2Θ 2(-g)

  32. PROJECTILE MOTION - SUMMARY • Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration • The projectile moves along a parabola

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