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Projectile motion
The document discusses projectile motion, providing equations and diagrams to analyze the trajectory, time, range, maximum height, and final velocity of objects projected either horizontally or at an angle. Key points covered include: - Horizontal projection results in a parabolic trajectory with the total time, horizontal range, and final velocity dependent on the initial height and velocity. - Angled projection also follows a parabolic trajectory, with the total time and horizontal range maximized at a launch angle of 45 degrees. The maximum height reached depends only on the initial velocity and launch angle.
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Projectile motion
- 1. PROJECTILE MOTION Senior HighSchool Physics 2010 Part 1. Free powerpoints at http://www.worldofteaching.com Part 2.
- 2. Introduction Projectile Motion: Motion throughthe air without a propulsion Examples:
- 3. Part 1. Motion ofObjects Projected Horizontally
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- 9. y •Motion is accelerated •Accelerationis constant, and downward • a = g = -9.81m/s2 g = -9.81m/s2 •The horizontal (x) component of velocity is constant •The horizontal and vertical motions are independent of each other, x but they have a common time
- 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. Frame of reference: Equationsof motion: y X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = -9.81 m/s2 VELC. vx = v0 vy = g t x DSPL. x = v0 t y = h + ½ g t2 v0 g h 0
- 12. Trajectory y x = v0t y = h + ½ g t2 Eliminate time, t t = x/v0 y = h + ½ g (x/v0)2 Parabola, open down h v01 v02 > v01 y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h x
- 13. Total Time, Δt y= h + ½ g t2 final y = 0 y 0 = h + ½ g (Δt)2 Δt = tf - ti ti =0 Solve for Δt: h Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms-2) Total time of motion depends only on the initial height, h tf =Δt x
- 14. Horizontal Range, Δx x= v0 t y final y = 0, time is the total time Δt Δx = v0 Δt h Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v0 Δx x
- 15. VELOCITY vx = v0 v= √vx 2 + vy 2 = √v02+g2t2 tg Θ = v / v = g t / v vy = g t Θ v
- 16. FINAL VELOCITY vx =v0 Δt = √ 2h/(-g) v = √vx 2 + vy 2 vy = g t v = √v02+g2(2h /(-g)) v = √ v02+ 2h(-g) Θ tg Θ = g Δt / v0 v = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 Θ is negative (below the horizontal line)
- 17. HORIZONTAL THROW -Summary h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2 Trajectory Half -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v0 √ 2h/(-g) Final Velocity v = √ v02+ 2h(-g) tg Θ = -√2h(-g) / v0
- 18. Part 2. Motion ofobjects projected at an angle
- 19. y vi Initial position: x= 0, y = 0 Initial velocity: vi = vi [Θ] viy Velocity components: x- direction : vix = vi cos Θ θ y- direction : viy = vi sin Θ x vix
- 20. y a =g= - 9.81m/s2 • Motionis 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: X Uniformmotion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.81 m/s2 VELOCITY vx = vix= vi cos Θ vy = viy+ g t vx = vi cos Θ vy = vi sin Θ + g t x = vix t = vi t cos Θ y = h + viy t + ½ g t2 x = vi t cos Θ y = vi t sin Θ + ½ g t2 DISPLACEMENT
- 23. Equations of motion: X Uniformmotion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.81 m/s2 VELOCITY vx = vi cos Θ vy = vi sin Θ + g t DISPLACEMENT x = vi t cos Θ y = vi t sin Θ + ½ g t2
- 24. x = vit cos Θ Trajectory y = vi t sin Θ + ½ g t2 y Parabola, open down Eliminate time, t t = x/(vi cos Θ) vi x sin Θ gx 2 y= + 2 vi cos Θ 2vi cos 2 Θ y = x tan Θ + g x2 2vi2 cos 2 Θ y = bx + ax2 x
- 25. Total Time, Δt y= vi t sin Θ + ½ g t2 final height y = 0, after time interval Δt 0 = vi Δt sin Θ + ½ g (Δt)2 Solve for Δt: x 0 = vi sin Θ + ½ g Δt Δt = 2 vi sin Θ (-g) t=0 Δt
- 26. Horizontal Range, Δx x= vi t cos Θ y final y = 0, time is the total time Δt Δx = vi Δt cos Θ Δt = Δx = 2 vi sin Θ (-g) sin (2 Θ) = 2 sin Θ cos Θ 2vi 2 sin Θ cos Θ (-g) x 0 Δx = Δx vi 2 sin (2 Θ) (-g)
- 27. Horizontal Range, Δx Δx= Θ (deg) sin (2 Θ) 0 0.00 15 0.50 30 1.00 60 0.87 75 0.50 90 0 (-g) •CONCLUSIONS: 0.87 45 vi 2 sin (2 Θ) •Horizontal range is greatest for the throw angle of 450 • Horizontal ranges are the same for angles Θ and (900 – Θ)
- 28. Trajectory and horizontalrange g y = x tan Θ + 2 x2 2vi cos 2 Θ 35 vi = 25 m/s 30 15 deg 30 deg 25 45 deg 20 60 deg 15 75 deg 10 5 0 0 20 40 60 80
- 29. Velocity •Final speed =initial speed (conservation of energy) •Impact angle = - launch angle (symmetry of parabola)
- 30. Maximum Height vy =vi sin Θ + g t y = vi t sin Θ + ½ g t2 At maximum height vy = 0 0 = vi sin Θ + g tup tup = vi sin Θ hmax = vi t upsin Θ + ½ g tup2 hmax = vi2 sin2 Θ/(-g) + ½ g(vi2 sin2 Θ)/g2 vi2 sin2 Θ (-g) tup = Δt/2 hmax = 2(-g)
- 31. Projectile Motion –Final Equations (0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 Trajectory Parabola, open down Total time Δt = Horizontal range Max height Δx = 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) hmax = vi2 sin2 Θ 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