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PROJECTILE MOTION Free powerpoints at http://www.worldofteaching.comhttp://www.worldofteaching.com

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Introduction Projectile Motion: Motion through the air without a propulsion Examples:

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Part 1. Motion of Objects Projected Horizontally

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v0v0 x y

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x y

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x y

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x y

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x y

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x y Motion is accelerated Acceleration is constant, and downward a = g = -9.81m/s 2 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/s 2

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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?

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x y 0 Frame of reference: h v0v0 Equations of motion: X Uniform m. Y Accel. m. ACCL.a x = 0a y = g = -9.81 m/s 2 VELC.v x = v 0 v y = g t DSPL.x = v 0 ty = h + ½ g t 2 g

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Trajectory x = v 0 t y = h + ½ g t 2 Eliminate time, t t = x/v 0 y = h + ½ g (x/v 0 ) 2 y = h + ½ (g/v 0 2 ) x 2 y = ½ (g/v 0 2 ) x 2 + h y x h Parabola, open down v 01 v 02 > v 01

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Total Time, Δt y = h + ½ g t 2 final y = 0 y x h t i =0 t f =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 = t f - t i

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

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VELOCITY v v x = v 0 v y = g t v = v x 2 + v y 2 = v 0 2 +g 2 t 2 tg Θ = v y / v x = g t / v 0 Θ

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FINAL VELOCITY v v x = v 0 v y = g t v = v x 2 + v y 2 v = v 0 2 +g 2 (2h /(-g)) v = v 0 2 + 2h(-g) Θ tg Θ = g Δt / v 0 = -(-g)2h/(-g) / v 0 = - 2h(-g) / v 0 Θ is negative (below the horizontal line) Δt = 2h/(-g)

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HORIZONTAL THROW - Summary TrajectoryHalf -parabola, open down Total time Δt = 2h/(-g) Horizontal Range Δx = v 0 2h/(-g) Final Velocity v = v 0 2 + 2h(-g) tg Θ = -2h(-g) / v 0 h – initial height, v 0 – initial horizontal velocity, g = -9.81m/s 2

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Part 2. Motion of objects projected at an angle

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vivi x y θ v ix v iy Initial velocity: vi = v i [Θ] Velocity components: x- direction : v ix = v i cos Θ y- direction : v iy = v i sin Θ Initial position: x = 0, y = 0

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x y Motion is accelerated Acceleration is constant, and downward a = g = -9.81m/s 2 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/s 2

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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?

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Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION a x = 0a y = g = -9.81 m/s 2 VELOCITY v x = v ix = v i cos Θ v x = v i cos Θ v y = v iy + g t v y = v i sin Θ + g t DISPLACEMENT x = v ix t = v i t cos Θ x = v i t cos Θ y = h + v iy t + ½ g t 2 y = v i t sin Θ + ½ g t 2

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Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION a x = 0a y = g = -9.81 m/s 2 VELOCITY v x = v i cos Θv y = v i sin Θ + g t DISPLACEMENT x = v i t cos Θy = v i t sin Θ + ½ g t 2

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Trajectory x = v i t cos Θ y = v i t sin Θ + ½ g t 2 Eliminate time, t t = x/(v i cos Θ) y x Parabola, open down y = bx + ax 2

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Total Time, Δt final height y = 0, after time interval Δt 0 = v i Δt sin Θ + ½ g (Δt) 2 Solve for Δt: y = v i t sin Θ + ½ g t 2 0 = v i sin Θ + ½ g Δt Δt = 2 v i sin Θ (-g) t = 0 Δt x

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Horizontal Range, Δx final y = 0, time is the total time Δt x = v i t cos Θ Δ x = v i Δ t cos Θ x Δx y 0 Δt = 2 v i sin Θ (-g) Δ x = 2v i 2 sin Θ cos Θ (-g) Δ x = v i 2 sin (2 Θ) (-g) sin (2 Θ) = 2 sin Θ cos Θ

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Horizontal Range, Δx Δ x = v i 2 sin (2 Θ) (-g) Θ (deg)sin (2 Θ) 00.00 150.50 300.87 451.00 600.87 750.50 900 CONCLUSIONS: Horizontal range is greatest for the throw angle of 45 0 Horizontal ranges are the same for angles Θ and (90 0 – Θ)

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Trajectory and horizontal range v i = 25 m/s

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

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Maximum Height v y = v i sin Θ + g t y = v i t sin Θ + ½ g t 2 At maximum height v y = 0 0 = v i sin Θ + g t up t up = v i sin Θ (-g) t up = Δt/2 h max = v i t up sin Θ + ½ g t up 2 h max = v i 2 sin 2 Θ/(-g) + ½ g(v i 2 sin 2 Θ)/g 2 h max = v i 2 sin 2 Θ 2(-g)

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Projectile Motion – Final Equations Trajectory Parabola, open down Total time Δt = Horizontal range Δx = Max height h max = (0,0) – initial position, v i = v i [Θ]– initial velocity, g = -9.81m/s 2 2 v i sin Θ (-g) v i 2 sin (2 Θ) (-g) v i 2 sin 2 Θ 2(-g)

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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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