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PROJECTILE MOTION Senior High School Physics

Study of object motion through the air without propulsion, including trajectory, velocity, range, and maximum height. Equations and analysis for both horizontal and angle projections. Includes free PowerPoint resources.

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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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