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Projectile Motion

Published in: Physics
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  • Debarun S

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This presentation clearly describes about projectile motion i.e object thrown at an angle, horizontally, verti cally, maximum height attained everything. hope it will help you

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    Part 1. PROJECTILE MOTION Part 2.
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    Introduction Projectile Motion: Motion through the air without a propulsion Examples:
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    Motion of Objects Projected Horizontally
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    О О О
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    О О О
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    0---0----0---0- --0---0- -O -O
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    0---0----0----0- --0---0---0 -O
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    O- -O- --0---0 --0 --0 --0---0 O '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 x
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    ANALYSIS OF MOTION ASSUMPTIONS: x-direction (horizontal): y-direction (vertical): no air resistance QUESTIONS: What is the trajectory? uniform motion accelerated motion What is the total time of the motion? What is the horizontal range? What is the final velocity?
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    Frame of reference: Equations of motion: Uniform m. Accele m, ay = g = -9.81 h ACCL VELC. x DSPL. a y I-n/s 2 = h + 1/2 g t2
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    Trajectory Y = h + 1/2 g t2 Eliminate time, t t = x/vo y = h + h g (x/vo)2 y = h + h (g/v02) x2 y = h (g/v02) x2 + h h Parabola, open down x
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    Total Time, At At Y = h + 1/2 g t2 final y = 0 O = h + 1/2 g (At)2 Solve for A t: 2h/(-g) At At = Al Ims-2) Total time of motion depends only on the initial height, h h tf =At x
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    Horizontal Range, Ax final y = 0, time is the total time At Ax = vo At At = Al 2h/(-g) Ax = vo 2h/(-g) h Ax orlzonta range epends on the initial height, h, and the initial x velocity, vo
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    VELOCITY vy = g t Nv02+g2t2 /v =gt/v
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    FINAL VELOCITY At 2h/(-g) 2 qv02+g2(2h /(-g)) V 02+ 2h(-g) tg @ = g At / vo = -(-g) 2h/(-g) / vo = -42h(-g) / vo @ is negative (below the horizontal line)
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    HORIZONTAL THROW - Summary h — initial height, vo — initial horizontal velocity, g = -9.81m/s2 Trajectory Total time Horizontal Range Final Velocity Half -parabola, open down At = Al 2h/(-g) Ax = vo 2h/(-g) v 02+ 2h(-g) tg @ = -42h(-g) / vo
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    Part 2, Motion of objects projected at an angle
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    0 Initial position: x = 0, y 0 Initial velocity: vi Velocity components: X- direction : Vix = Vi COS @ y- direction : Viy = Vi sin @ x
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    o oo O O - 9.81m/s2 O O 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 O o x
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    ANALYSIS OF MOTION: ASSUMPTIONS x-direction (horizontal): y-direction (vertical): no air resistance QUESTIONS What is the trajectory? uniform motion accelerated motion What is the total time of the motion? What is the horizontal range? What is the maximum height? What is the final velocity?
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    ACCELERATION VELOCITY DISPLACEMENT Equations of motion: Uniform motion = Vix= Vi COS @ v = COS @ v X = Vikt = t COS @ x = t cos @ Accelerated motion ay = g = -9.81 m/s2 = Vi Sin @ + g t V = h + Viy t + 1/2 g t2 y y = t sin @ + h g t2
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    Accelerated motion ACCELERATION VELOCITY DISPLACEMENT Equations of motion: Uniform motion x = COS @ v x = t cos @ ay = g = -9.81 m/s2 = Vi Sin @ + g t V y = t sin @ + h g t2
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    Trajectory x = t cos @ Y = t sin @ + h g t2 Eliminate time, t t = x/(Vi cos O) Vixsin @ COS@ 2Vi coe @ Y = x tan @ -k 2v: coe @ 2 x + ax Parabola, open down x
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    Total Time, At Y = t sin @ + h g t2 final height y = 0, after time interval At 0 = ViAt sin @ + h g (At)2 Solve for A t: 0 = Vi sin @ + h g At 2 Vi sin @ At = x At
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    Horizontal Range, Ax x = t COS @ final y = 0, time is the total time At Ax = At cos @ 2 Vi sin @ At sin (2 Q) 2Vi 2 sin @ cos @ Ax = 2 sin @ cos @ x Ax Vi2 sin (2 0) Ax =
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    Horizontal Range, Ax Vi2 sin (2 0) Ax = @ (deg) sin (2 Q) 15 30 45 60 75 90 0.00 0.50 0.87 1.00 0.87 0.50 •CONCLUSIONS: 'Horizontal range is greatest for the throw angle of 450 ' Horizontal ranges are the same for 900 — @ angles @ and (
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    Trajectory and horizontal range Y = xtan@+ 2Vi COS2 @ 35 30 25 20 15 10 20 v. = 25 m/s 40 — 15 deg — 30 deg — 45 deg — 60 deg — 75 deg 60 80
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    Velocity 'Final speed = initial speed (conservation of energy) 'Impact angle = - launch angle (symmetry of parabola)
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    Maximum Height v = sin @ + gt Y = Vi t sin @ + h g t2 At maximum heightv = 0 0 = Vi sin @ + g tup hmax = tupsin @ + h g tup Vi sin @ tup - = At/2 tup h hmax 2 = v.2 sin2 @/(-g) + h g(Vi2 sin2 Vi2 sin2 @ 2(-g)
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    Projectile Motion - Final Equations (0,0) — initial position, Vi = Vi initial velocity, g = -9.81m/s2 Trajectory Total time Parabola, open down 2 Vi sin @ At 2 sin (2 0) Ax Horizontal range Max height Vi2 sin2 @ hmax 2(-g)
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    PROJECTILE MOTION - SUMMARY Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration The projectile moves along a parabola Member Only AnimationFactorv com

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