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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.comIntroductionProjectile Motion: Motion through the air without a propulsionExamples: Part 1.Motion of Objects Projected Horizontallyyv0xyxyxyxyxyMotion 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 MOTIONASSUMPTIONS: 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/s2Motion 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, ΔxCONCLUSIONS: 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/sVelocityFinal 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 - SUMMARYProjectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration The projectile moves along a parabola

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