PROJECTILE MOTION Senior High School Physics Lech Jedral

PROJECTILE MOTION Senior High School Physics Lech Jedral 2006 Part 1. Part 2.

Introduction v Projectile Motion: Motion through the air without a propulsion v Examples:

Part 1. Motion of Objects Projected Horizontally

y v 0 x

y x

y x

y x

y x

y • Motion is accelerated • Acceleration is constant, and downward • a = g = -9. 81 m/s 2 • The horizontal (x) component of velocity is constant • The horizontal and vertical motions are independent of each other, but they have a x common time

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?

Frame of reference: y v 0 g h 0 Equations of motion: X Y Uniform m. Accel. m. ACCL. ax = 0 ay = g = -9. 81 m/s 2 VELC. vx = v 0 vy = g t x DSPL. x = v 0 t y = h + ½ g t 2

Trajectory y x = v 0 t y = h + ½ g t 2 Eliminate time, t t = x/v 0 y = h + ½ g (x/v 0)2 Parabola, open down h v 01 v 02 > v 01 y = h + ½ (g/v 02) x 2 y = ½ (g/v 02) x 2 + h x

Total Time, Δt y = h + ½ g t 2 final y = 0 y 0 = h + ½ g (Δt)2 ti =0 Solve for Δt: h Δt = tf - ti Δt = √ 2 h/(-g) Δt = √ 2 h/(9. 81 ms-2) Total time of motion depends only on the initial height, h tf =Δt x

Horizontal Range, Δx x = v 0 t y final y = 0, time is the total time Δt Δx = v 0 Δt h Δt = √ 2 h/(-g) Δx = v 0 √ 2 h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v 0 Δx x

VELOCITY vx = v 0 Θ v = √vx 2 + vy 2 = √v 02+g 2 t 2 tg Θ = v / v = g t / v y x 0 vy = g t v

FINAL VELOCITY vx = v 0 Δt = √ 2 h/(-g) v = √vx 2 + vy 2 vy = g t v = √v 02+g 2(2 h /(-g)) v=√ v 02+ 2 h(-g) tg Θ = g Δt / v 0 Θ v = -(-g)√ 2 h/(-g) / v 0 = -√ 2 h(-g) / v 0 Θ is negative (below the horizontal line)

HORIZONTAL THROW - Summary h – initial height, v 0 – initial horizontal velocity, g = -9. 81 m/s 2 Trajectory Half -parabola, open down Total time Δt = √ 2 h/(-g) Horizontal Range Δx = v 0 √ 2 h/(-g) Final Velocity v = √ v 02+ 2 h(-g) tg Θ = -√ 2 h(-g) / v 0

Part 2. Motion of objects projected at an angle

y vi Initial position: x = 0, y = 0 Initial velocity: vi = vi [Θ] viy Velocity components: x- direction : vix = vi cos Θ θ y- direction : viy = vi sin Θ x vix

y a =g= - 9. 81 m/s 2 • Motion is accelerated • Acceleration is constant, and downward • a = g = -9. 81 m/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 x

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?

Equations of motion: X Uniform motion ax = 0 Y Accelerated motion ay = g = -9. 81 m/s 2 VELOCITY vx = vix= vi cos Θ vx = vi cos Θ vy = viy+ g t vy = vi sin Θ + g t DISPLACEMENT x = vix t = vi t cos Θ y = h + viy t + ½ g t 2 x = vi t cos Θ y = vi t sin Θ + ½ g t 2 ACCELERATION

Equations of motion: X Uniform motion ax = 0 Y Accelerated motion ay = g = -9. 81 m/s 2 VELOCITY vx = vi cos Θ vy = vi sin Θ + g t DISPLACEMENT x = vi t cos Θ y = vi t sin Θ + ½ g t 2 ACCELERATION

Trajectory x = vi t cos Θ y = vi t sin Θ + ½ g t 2 y Parabola, open down Eliminate time, t t = x/(vi cos Θ) y = bx + ax 2 x

Total Time, Δt y = vi t sin Θ + ½ g t 2 final height y = 0, after time interval Δt 0 = vi Δt sin Θ + ½ g (Δt)2 Solve for Δt: x 0 = vi sin Θ + ½ g Δt Δt = 2 vi sin Θ (-g) t=0 Δt

Horizontal Range, Δx x = vi t cos Θ y final y = 0, time is the total time Δt Δx = vi Δt cos Θ Δt = Δx = 2 vi sin Θ (-g) sin (2 Θ) = 2 sin Θ cos Θ 2 vi 2 sin Θ cos Θ (-g) x 0 Δx = Δx vi 2 sin (2 Θ) (-g)

Horizontal Range, Δx Δx = Θ (deg) sin (2 Θ) 0 0. 00 15 0. 50 30 0. 87 45 1. 00 60 0. 87 75 0. 50 90 0 vi 2 sin (2 Θ) (-g) • 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 range vi = 25 m/s

Velocity • Final speed = initial speed (conservation of energy) • Impact angle = - launch angle (symmetry of parabola)

Maximum Height vy = vi sin Θ + g t y = vi t sin Θ + ½ g t 2 At maximum height vy = 0 0 = vi sin Θ + g tup = vi sin Θ (-g) tup = Δt/2 hmax = vi t upsin Θ + ½ g tup 2 hmax = vi 2 sin 2 Θ/(-g) + ½ g(vi 2 sin 2 Θ)/g 2 hmax = vi 2 sin 2 Θ 2(-g)

Projectile Motion – Final Equations (0, 0) – initial position, vi = vi [Θ]– initial velocity, g = -9. 81 m/s 2 Trajectory Total time Horizontal range Parabola, open down Δt = Δx = 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) vi 2 sin 2 Θ Max height hmax = 2(-g)

PROJECTILE MOTION - SUMMARY v Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration v The projectile moves along a parabola