PROJECTILEMOTIONChapter1.4.ppt

PROJECTILE MOTIONSenior High School Physics Lech Jedral 2006 Introduction Projectile Motion: Motion through the air without a propulsion Examples: Part 1.Motion of Objects Projected Horizontally Trajectory Total Time, Δt Horizontal Range, Δx VELOCITY FINAL VELOCITY HORIZONTAL THROW - Summary Part 2.Motion of objects projected at an angle Trajectory Total Time, Δt Horizontal Range, Δx Horizontal Range, Δx Trajectory and horizontal range Velocity Maximum Height Projectile Motion – Final Equations PROJECTILE MOTION - SUMMARY Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration The projectile moves along a parabola hmax = Max height Δx = Horizontal range Δt = Total time Parabola, open down Trajectory (0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) vi2 sin2 Θ 2(-g) * * Part 1. Part 2. Free powerpoints at v0 x y x y x y x y x y 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 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? x y 0 Frame of reference: h v0 Equations of motion: y = h + ? g t2 x = v0 t DSPL. vy = g t vx = v0 VELC. ay = g = -9.81 m/s2 ax = 0 ACCL. Y Accel. m. X Uniform m. g x = v0 t y = h + ? g t2 Eliminate time, t t = x/v0 y = h + ? g (x/v0)2 y = h + ? (g/v02) x2 y = ? (g/v02) x2 + h y x h Parabola, open down v01 v02 v01 y = h + ? g t2 final y = 0 y x h ti =0 tf =Δt 0 = h + ? 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 -