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PROJECTILE MOTION Free powerpoints at

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**Introduction Projectile Motion:**

Motion through the air without a propulsion Examples:

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**Part 1. Motion of Objects Projected Horizontally**

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y v0 x

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y x

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y x

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y x

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y x

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**g = -9.81m/s2 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 x

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**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?

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**X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = -9.81 m/s2 VELC.**

Frame of reference: Equations of motion: x y X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = m/s2 VELC. vx = v0 vy = g t DSPL. x = v0 t y = h + ½ g t2 v0 h g

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**Trajectory y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h x = v0 t**

y = h + ½ g t2 Parabola, open down Eliminate time, t v02 > v01 v01 t = x/v0 y = h + ½ g (x/v0)2 y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h

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**Total Time, Δt Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms-2) y = h + ½ g t2**

Δt = tf - ti y = h + ½ g t2 y x h final y = 0 0 = h + ½ g (Δt)2 ti =0 Solve for Δt: Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms-2) tf =Δt Total time of motion depends only on the initial height, h

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**Horizontal Range, Δx Δx = v0 Δt Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δx**

y x h final y = 0, time is the total time Δt Δx = v0 Δt Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δx Horizontal range depends on the initial height, h, and the initial velocity, v0

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**VELOCITY Θ v = √v02+g2t2 tg Θ = vy/ vx = g t / v0 vx = v0 vy = g t**

v = √vx2 + vy2 = √v02+g2t2 tg Θ = vy/ vx = g t / v0

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**FINAL VELOCITY Δt = √ 2h/(-g) Θ v v = √v02+g2(2h /(-g))**

vx = v0 Δt = √ 2h/(-g) v tg Θ = g Δt / v0 = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 Θ vy = g t v = √vx2 + vy2 v = √v02+g2(2h /(-g)) v = √ v02+ 2h(-g) Θ is negative (below the horizontal line)

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**HORIZONTAL THROW - Summary**

h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2 Trajectory Half -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v0 √ 2h/(-g) Final Velocity v = √ v02+ 2h(-g) tg Θ = -√2h(-g) / v0

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**Part 2. Motion of objects projected at an angle**

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

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**y a = g = - 9.81m/s2 x 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): 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?

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**X Uniform motion Y Accelerated motion**

Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = m/s2 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 x = vi t cos Θ y = h + viy t + ½ g t2 y = vi t sin Θ + ½ g t2

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**X Uniform motion Y Accelerated motion**

Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = m/s2 VELOCITY vx = vi cos Θ vy = vi sin Θ + g t DISPLACEMENT x = vi t cos Θ y = vi t sin Θ + ½ g t2

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**Trajectory x = vi t cos Θ y = vi t sin Θ + ½ g t2 Parabola, open down**

Eliminate time, t t = x/(vi cos Θ) y = bx + ax2 x

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**Total Time, Δt y = vi t sin Θ + ½ g t2 Δt =**

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

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**Horizontal Range, Δx Δx = vi Δt cos Θ Δt = Δx = Δx = x = vi t cos Θ Δx**

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

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**Horizontal Range, Δx Δx = vi 2 sin (2 Θ) Θ (deg) sin (2 Θ) (-g)**

0.00 15 0.50 30 0.87 45 1.00 60 75 90 CONCLUSIONS: Horizontal range is greatest for the throw angle of 450 Horizontal ranges are the same for angles Θ and (900 – Θ)

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**Trajectory and horizontal range**

vi = 25 m/s

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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 vy = vi sin Θ + g t y = vi t sin Θ + ½ g t2**

At maximum height vy = 0 0 = vi sin Θ + g tup hmax = vi t upsin Θ + ½ g tup2 hmax = vi2 sin2 Θ/(-g) + ½ g(vi2 sin2 Θ)/g2 hmax = vi2 sin2 Θ 2(-g) tup = vi sin Θ (-g) tup = Δt/2

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**Projectile Motion – Final Equations**

(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2 Trajectory Parabola, open down Total time Δt = Horizontal range Δx = Max height hmax = 2 vi sin Θ (-g) vi 2 sin (2 Θ) (-g) vi2 sin2 Θ 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

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