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PROJECTILE MOTION Free powerpoints at http://www.worldofteaching.comhttp://www.worldofteaching.com Slide 2 Introduction Projectile Motion: Motion through the air without a propulsion Examples: Slide 3 Part 1. Motion of Objects Projected Horizontally Slide 4 v0v0 x y Slide 5 x y Slide 6 x y Slide 7 x y Slide 8 x y Slide 9 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 Slide 10 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? Slide 11 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 + g t 2 g Slide 12 Trajectory x = v 0 t y = h + g t 2 Eliminate time, t t = x/v 0 y = h + g (x/v 0 ) 2 y = h + (g/v 0 2 ) x 2 y = (g/v 0 2 ) x 2 + h y x h Parabola, open down v 01 v 02 > v 01 Slide 13 Total Time, t y = h + g t 2 final y = 0 y x h t i =0 t f =t 0 = h + g (t) 2 Solve for t: t = 2h/(-g) t = 2h/(9.81ms -2 ) Total time of motion depends only on the initial height, h t = t f - t i Slide 14 Horizontal Range, x final y = 0, time is the total time t y x h t = 2h/(-g) x = v 0 2h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v 0 x x = v 0 t x = v 0 t Slide 15 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 = v y / v x = g t / v 0 Slide 16 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) tg = g t / v 0 = -(-g)2h/(-g) / v 0 = - 2h(-g) / v 0 is negative (below the horizontal line) t = 2h/(-g) Slide 17 HORIZONTAL THROW - Summary TrajectoryHalf -parabola, open down Total time t = 2h/(-g) Horizontal Range x = v 0 2h/(-g) Final Velocity v = v 0 2 + 2h(-g) tg = -2h(-g) / v 0 h initial height, v 0 initial horizontal velocity, g = -9.81m/s 2 Slide 18 Part 2. Motion of objects projected at an angle Slide 19 vivi x y v ix v iy Initial velocity: vi = v i [] Velocity components: x- direction : v ix = v i cos y- direction : v iy = v i sin Initial position: x = 0, y = 0 Slide 20 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 Slide 21 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? Slide 22 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 v x = v i cos v y = v iy + g t v y = v i sin + g t DISPLACEMENT x = v ix t = v i t cos x = v i t cos y = h + v iy t + g t 2 y = v i t sin + g t 2 Slide 23 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 v y = v i sin + g t DISPLACEMENT x = v i t cos y = v i t sin + g t 2 Slide 24 Trajectory x = v i t cos y = v i t sin + g t 2 Eliminate time, t t = x/(v i cos ) y x Parabola, open down y = bx + ax 2 Slide 25 Total Time, t final height y = 0, after time interval t 0 = v i t sin + g (t) 2 Solve for t: y = v i t sin + g t 2 0 = v i sin + g t t = 2 v i sin (-g) t = 0 t x Slide 26 Horizontal Range, x final y = 0, time is the total time t x = v i t cos x = v i t cos x x y 0 t = 2 v i sin (-g) x = 2v i 2 sin cos (-g) x = v i 2 sin (2 ) (-g) sin (2 ) = 2 sin cos Slide 27 Horizontal Range, x x = v i 2 sin (2 ) (-g) (deg)sin (2 ) 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 and (90 0 ) Slide 28 Trajectory and horizontal range v i = 25 m/s Slide 29 Velocity Final speed = initial speed (conservation of energy) Impact angle = - launch angle (symmetry of parabola) Slide 30 Maximum Height v y = v i sin + g t y = v i t sin + g t 2 At maximum height v y = 0 0 = v i sin + g t up t up = v i sin (-g) t up = t/2 h max = v i t up sin + g t up 2 h max = v i 2 sin 2 /(-g) + g(v i 2 sin 2 )/g 2 h max = v i 2 sin 2 2(-g) Slide 31 Projectile Motion Final Equations Trajectory Parabola, open down Total time t = Horizontal range x = Max height h max = (0,0) initial position, v i = v i [] initial velocity, g = -9.81m/s 2 2 v i sin (-g) v i 2 sin (2 ) (-g) v i 2 sin 2 2(-g) Slide 32 PROJECTILE MOTION - SUMMARY Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration The projectile moves along a parabola