PROJECTILE MOTION Senior High School Physics

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PROJECTILE MOTION Senior High School Physics. Lech Jedral 2006. Part 1. Part 2. Free powerpoints at http://www.worldofteaching.com. Introduction. Projectile Motion: Motion through the air without a propulsion Examples:. Part 1. Motion of Objects Projected Horizontally. y. v 0. x. y.
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PROJECTILE MOTIONSenior High School PhysicsLech Jedral2006Part 1.Part 2.Free powerpoints at http://www.worldofteaching.comIntroduction
  • Projectile Motion:
  • Motion through the air without a propulsion
  • Examples:
  • Part 1.Motion of Objects Projected Horizontallyyv0xyxyxyxyxy
  • 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/s2xANALYSIS 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?
  • yv0ghx0Frame of reference:Equations of motion:yhv02 > v01v01xTrajectoryx = v0 ty = h + ½ g t2Parabola, open downEliminate time, tt = x/v0y = h + ½ g (x/v0)2y = h + ½ (g/v02) x2y = ½ (g/v02) x2 + hyhΔt = √ 2h/(9.81ms-2)Δt = √ 2h/(-g)xTotal Time, ΔtΔt = tf - tiy = h + ½ g t2final y = 00 = h + ½ g (Δt)2ti =0Solve for Δt:tf =ΔtTotal time of motion depends only on the initial height, h yhΔt = √ 2h/(-g)Δx = v0 √ 2h/(-g)ΔxxHorizontal Range, Δxx = v0 tfinal y = 0, time is the total time ΔtΔx = v0 ΔtHorizontal range depends on the initial height, h, and the initial velocity, v0vv = √vx2 + vy2= √v02+g2t2tg Θ = vy/ vx = g t / v0VELOCITYvx = v0Θvy = g tΔt = √ 2h/(-g)tg Θ = g Δt / v0= -(-g)√2h/(-g) / v0= -√2h(-g) / v0vv = √vx2 + vy2v = √v02+g2(2h /(-g))v = √ v02+ 2h(-g)FINAL VELOCITYvx = v0Θvy = g tΘ is negative (below the horizontal line)HORIZONTAL THROW - Summaryh – initial height, v0 – initial horizontal velocity, g = -9.81m/s2Part 2.Motion of objects projected at an angleyInitial velocity: vi = vi [Θ]Velocity components:x- direction : vix = vi cos Θy- direction : viy = vi sin ΘviyviθxvixInitial position: x = 0, y = 0ya = 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
  • xANALYSIS 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:Equations of motion:Trajectoryx = vi t cos Θy = vi tsin Θ + ½ g t2Parabola, open downyEliminate time, tt = x/(vi cos Θ)y = bx + ax2x0 =vi sin Θ + ½ g ΔtΔt = 2 vi sinΘ(-g)Total Time, Δty = vi tsin Θ + ½ g t2final height y = 0, after time interval Δt0 =vi Δtsin Θ + ½ g (Δt)2Solve for Δt:xt = 0 Δt2 vi sinΘΔt =(-g)Δx =2vi 2 sin Θ cos Θvi 2 sin (2Θ)Δx =(-g)(-g)Horizontal Range, Δxx = vi t cos Θyfinal y = 0, time is the total time ΔtΔx = vi Δt cos Θx0sin (2Θ) = 2 sin Θ cos ΘΔxvi 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)
  • Trajectory and horizontal rangevi = 25 m/sVelocity
  • Final speed = initial speed (conservation of energy)
  • Impact angle = - launch angle (symmetry of parabola)
  • tup=hmax =vi t upsin Θ + ½ g tup2hmax =vi2sin2Θ/(-g) + ½ g(vi2sin2Θ)/g2hmax =vi sin Θ(-g)vi2sin2Θ2(-g)Maximum Heightvy = vi sin Θ + g ty = vi tsin Θ + ½ g t2At maximum height vy = 00 = vi sin Θ + g tuptup= Δt/2vi 2 sin (2Θ)(-g)Projectile Motion – Final Equations(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s22 vi sinΘ(-g)vi2sin2Θ2(-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
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