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# Projectile motion (2)

## on Nov 01, 2011

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plus one projectile motion

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## Projectile motion (2)Presentation Transcript

• PROJECTILE MOTION
• Introduction
• Projectile Motion:
• Motion through the air without a propulsion
• Examples:
• Part 1. Motion of Objects Projected Horizontally
• v 0 x y
• x y
• x y
• x y
• x y
• 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
• 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?
• Frame of reference: Equations of motion: x y 0 h v 0 y = h + ½ g t 2 x = v 0 t DSPL. v y = g t v x = v 0 VELC. a y = g = -9.81 m/s 2 a x = 0 ACCL. Y Accel. m. X Uniform m. g
• 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 Parabola, open down y x h v 01 v 02 > v 01
• Total Time, Δt y = h + ½ g t 2 final y = 0 t i =0 t f = Δt 0 = h + ½ g ( Δt) 2 Solve for Δt: Total time of motion depends only on the initial height, h Δt = t f - t i y x h Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms -2 )
• Horizontal Range, Δx final y = 0, time is the total time Δt Horizontal range depends on the initial height, h, and the initial velocity, v 0 x = v 0 t Δ x = v 0 Δ t y x h Δt = √ 2h/(-g) Δx = v 0 √ 2h/(-g) Δx
• VELOCITY v x = v 0 v y = g t Θ v v = √v x 2 + v y 2 = √v 0 2 +g 2 t 2 tg Θ = v y / v x = g t / v 0
• FINAL VELOCITY v x = v 0 v y = g t Θ Θ is negative ( below the horizontal line) v v = √v x 2 + v y 2 v = √v 0 2 +g 2 (2h /(-g)) v = √ v 0 2 + 2h(-g) tg Θ = g Δt / v 0 = -(-g)√2h/(-g) / v 0 = - √2h(-g) / v 0 Δt = √ 2h/(-g)
• HORIZONTAL THROW - Summary h – initial height, v 0 – initial horizontal velocity, g = -9.81m/s 2 v = √ v 0 2 + 2h(-g) tg Θ = -√2h(-g) / v 0 Final Velocity Δx = v 0 √ 2h/(-g) Horizontal Range Δt = √ 2h/(-g) Total time Half -parabola, open down Trajectory
• Part 2. Motion of objects projected at an angle
• Initial position: x = 0, y = 0 v i x y θ v ix v iy Initial velocity: vi = v i [ Θ] Velocity components: x- direction : v 0x = v 0 cos Θ y- direction : v 0y = v 0 sin Θ
• 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
• 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?
• Equations of motion: y = v 0 t sin Θ + ½ g t 2 x = v 0x t = v 0 t cos Θ x = v 0 t cos Θ DISPLACEMENT v y = v 0 sin Θ - g t v x = v 0x = v 0 cos Θ VELOCITY a y = g = -9.81 m/s 2 a x = 0 ACCELERATION Y Accelerated motion X Uniform motion
• 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 = b x + a x 2
• Total Time, Δt final height y = 0, after time interval Δt 0 = v 0 Δ t sin Θ + ½ g ( Δ t) 2 Solve for Δt: y = v 0 t sin Θ + ½ g t 2 t = 0 Δ t x 0 = v 0 sin Θ + ½ g Δ t Δ t = 2 v 0 sin Θ (-g)
• Horizontal Range, Δx final y = 0, time is the total time Δt x = v 0 t cos Θ Δ x = v o Δ t cos Θ x Δx y 0 sin (2 Θ) = 2 sin Θ cos Θ Δt = 2 v 0 sin Θ (-g) Δ x = 2v o 2 sin Θ cos Θ (-g) Δ x = v o 2 sin (2 Θ) (-g)
• Horizontal Range, Δx
• CONCLUSIONS:
• Horizontal range is greatest for the throw angle of 45 0
• Horizontal ranges are the same for angles Θ and ( 90 0 – Θ )
Δ x = v 0 2 sin (2 Θ) (-g) 0.50 75 0.00 0 0 90 0.87 60 1.00 45 0.87 30 0.50 15 sin (2 Θ) Θ (deg)
• Trajectory and horizontal range v i = 25 m/s
• Velocity
• Final speed = initial speed (conservation of energy)
• Impact angle = - launch angle (symmetry of parabola)
• Maximum Height v y = v o sin Θ + g t y = v 0 t sin Θ + ½ g t 2 At maximum height v y = 0 0 = v i0 sin Θ + g t up t up = Δt/2 t up = v 0 sin Θ (-g) h max = v 0 t up sin Θ + ½ g t up 2 h max = v 0 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)
• Projectile Motion – Final Equations (0,0) – initial position, v i = v i [ Θ] – initial velocity, g = -9.81m/s 2 2 v 0 sin Θ (-g) v 0 2 sin 2 Θ 2(-g) h max = Max height Δx = Horizontal range Δt = Total time Parabola, open down Trajectory v 0 2 sin (2 Θ) (-g)
• PROJECTILE MOTION - SUMMARY
• Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration
• The projectile moves along a parabola