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### Projectile motion

• 2. Introduction  Projectile Motion: Motion through the air without a propulsion  Examples:
• 3. Part 1. Motion of Objects Projected Horizontally
• 5. x y
• 6. x y
• 7. x y
• 8. x y
• 9. x 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
• 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. x y 0 Frame of reference: h v0 Equations of motion: X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = -9.81 m/s2 VELC. vx = v0 vy = g t DSPL. x = v0 t y = h + ½ g t2 g
• 12. Trajectory x = v0 t y = h + ½ g t2 Eliminate time, t t = x/v0 y = h + ½ g (x/v0)2 y = h + ½ (g/v0 2 ) x2 y = ½ (g/v0 2 ) x2 + h y x h Parabola, open down v01 v02 > v01
• 13. Total Time, Δt y = h + ½ g t2 final y = 0 y x h ti =0 tf =Δt 0 = h + ½ g (Δt)2 Solve for Δt: Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms-2 ) Total time of motion depends only on the initial height, h Δt = tf - ti
• 14. Horizontal Range, Δx final y = 0, time is the total time Δt y x h Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v0 Δx x = v0 t Δx = v0 Δt
• 15. VELOCITY v vx = v0 vy = g t v = √vx 2 + vy 2 = √v0 2 +g2 t2 tg Θ = v / v = g t / v Θ
• 16. FINAL VELOCITY v vx = v0 vy = g t v = √vx 2 + vy 2 v = √v0 2 +g2 (2h /(-g)) v = √ v0 2 + 2h(-g) Θ tg Θ = g Δt / v0 = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 Θ is negative (below the horizontal line) Δt = √ 2h/(-g)
• 17. HORIZONTAL THROW - Summary Trajectory Half -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v0 √ 2h/(-g) Final Velocity v = √ v0 2 + 2h(-g) tg Θ = -√2h(-g) / v0 h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2
• 18. Part 2. Motion of objects projected at an angle
• 19. vi x y θ vix viy Initial velocity: vi = vi [Θ] Velocity components: x- direction : vix = vi cos Θ y- direction : viy = vi sin Θ Initial position: x = 0, y = 0
• 20. x 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 a = g = - 9.81m/s2
• 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 Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.81 m/s2 VELOCITY vx = vix= vi cos Θ vx = vi cos Θ vy = viy+ g t vy = vi sin Θ + g t DISPLACEMENT x = vix t = vi t cos Θ x = vi t cos Θ y = h + viy t + ½ g t2 y = vi tsin Θ + ½ g t2
• 23. Equations of motion: X Uniform motion 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 tsin Θ + ½ g t2
• 24. Trajectory x = vi t cos Θ y = vi tsin Θ + ½ g t2 Eliminate time, t t = x/(vi cos Θ) y x Parabola, open down 2 22 22 2 cos2 tan cos2cos sin x v g xy v gx v xv y i ii i Θ +Θ= Θ + Θ Θ = y = bx + ax2
• 25. Total Time, Δt final height y = 0, after time interval Δt 0 = vi Δtsin Θ + ½ g (Δt)2 Solve for Δt: y = vi tsin Θ + ½ g t2 0 = vi sin Θ + ½ g Δt Δt = 2 vi sin Θ (-g) t = 0 Δt x
• 26. Horizontal Range, Δx final y = 0, time is the total time Δt x = vi t cos Θ Δx = vi Δt cos Θ x Δx y 0 Δt = 2 vi sin Θ (-g) Δx = 2vi 2 sin Θ cos Θ (-g) Δx = vi 2 sin (2 Θ) (-g) sin (2 Θ) = 2 sin Θ cos Θ
• 27. Horizontal Range, Δx Δx = vi 2 sin (2 Θ) (-g) Θ (deg) sin (2Θ) 0 0.00 15 0.50 30 0.87 45 1.00 60 0.87 75 0.50 90 0 •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 2 22 cos2 tan x v g xy i Θ +Θ= 0 5 10 15 20 25 30 35 0 20 40 60 80 15 deg 30 deg 45 deg 60 deg 75 deg vi = 25 m/s
• 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 tsin Θ + ½ g t2 At maximum height vy = 0 0 = vi sin Θ + g tup tup = vi sin Θ (-g) tup = Δt/2 hmax = vi tupsin Θ + ½ g tup 2 hmax = vi 2 sin2 Θ/(-g) + ½ g(vi 2 sin2 Θ)/g2 hmax = vi 2 sin2 Θ 2(-g)
• 31. Projectile Motion – Final Equations Trajectory Parabola, open down Total time Δt = Horizontal range Δx = Max height hmax = (0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) vi 2 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
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