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Trajectory Equation

Trajectory Equation. With the cubbies!. Title. Text Subtext. Introduction. Projectile Motion: Motion through the air without a propulsion Examples:. Part 1. Motion of Objects Projected Horizontally. y. v 0. x. y. x. y. x. y. x. y. x. y. Motion is accelerated

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Trajectory Equation

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  1. Trajectory Equation With the cubbies!

  2. Title • Text • Subtext

  3. Introduction • Projectile Motion: Motion through the air without a propulsion • Examples:

  4. Part 1.Motion of Objects Projected Horizontally

  5. y v0 x

  6. y x

  7. y x

  8. y x

  9. y x

  10. 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

  11. 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?

  12. y v0 g h x 0 Frame of reference: Equations of motion:

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

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

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

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

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

  18. HORIZONTAL THROW - Summary h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2

  19. Trajectory Equation • Let's derive the equation of the trajectory of a free-falling object. • Let's set up the problem by assuming that the object has an initial velocity of "v0" directed at "θ" to the horizontal.

  20. Grrrrr.

  21. Trajectory Motion • We will make the math simpler by defining the origin of our coordinate system to coincide with the object's initial location. • We can, therefore, write down its horizontal coordinate as a function of time.

  22. Trajectory Equation • We can rearrange this equation and insert "v0cosθ" for "vxi" which results in

  23. Trajectory Equation • Now let's turn our attention to the vertical coordinate. Its equation as a function of time is

  24. Trajectory Equation • We can insert "v0·sinθ" for "vyi" and eliminate "t" from this equation by using the equation we obtained for the horizontal coordinate. This leads to

  25. Trajectory Equation • This equation has the form Which we recognize as a parabola

  26. Where should the hunter aim to hit the monkey?

  27. Using Components in 2 dimensional projectile motion • A projectile is any object that’s motion is influenced by gravity

  28. Using Components in 2 dimensional projectile motion • The velocity vector is drawn tangent to the path. Vi

  29. Using Components in 2 dimensional projectile motion • Horizontal and vertical components can be drawn for each vector. Vi

  30. We can separate the motion into two problems a. horizontal b Vertical • We can separate the motion into two problem . Vi

  31. The horizontal component • There is no acceleration in the x direction so the horizontal velocity is constant in the horizontal direction. Xf – Xi = Vx*t + 0 Vix = Vi cos0 Vfx = Vfi Vfx = Vf cos0

  32. Solving the vertical problem • We will use the same equations we used in chapter for except we will use y in place of d. Yf Yi yf- yi = Vyt + 1/2gt2 Vyf= Vfsin0 Vyf = Vi + gt Vyi = ViSin0 Vyf2 – Vyi2 = 2g(yf –yi) Vi

  33. 1. A stone is thrown horizontally at a speed of +5.0 m/s from the top of a cliff 78.4 m high • A. How long does it take the stone to reach the bottom of the cliff? Unknown t Given yf = 78.4m yi = 0m Vyi = 0m/s Vxi = 5 m/s g= -9.8m/s Vx= 5m/s Equation yfy –yiy = Viyt +1/2gt2 78.4m Solving for t yf = 1/2gt2 t2 = 2yf/g t = 2yf/g = 2*78.4m/9.8m/s2 t = 4 s

  34. 1. A stone is thrown horizontally at a speed of +5.0 m/s from the top of a cliff 78.4 m high • B. How far from the base of the cliff does the stone land? Unknown x Given yf = 78.4m yi = 0m Vyi = 0m/s Vxi = 5 m/s g= -9.8m/s Vx= 5m/s Equation X = Vx*t 78.4m Solving for x X = 5m/s * 4.00s X = 20 m

  35. 1. A stone is thrown horizontally at a speed of +5.0 m/s from the top of a cliff 78.4 m high Unknown Vfx, Vfy Given yf = 78.4m yi = 0m Vyi = 0m/s Vxi = 5 m/s g= -9.8m/s t =4s • C. What are the horizontal and vertical components of the velocity just before it hits the ground? Equation for Vxf Vxf = Vxi Vxf = 5m/s Vx= 5m/s Solving for Vyf Vyf = Vif + gt Vyf = 0 + gt Vyf = 9.8m/s2 * 4s Vyf = 39.2 m/s 78.4m Vf

  36. 1. Consider these diagrams in answering the following questions. • Which diagram (if any) might represent ... • a. ... the initial horizontal velocity? • b. ... the initial vertical velocity? • c. ... the horizontal acceleration? • d. ... the vertical acceleration? • e. ... the net force?

  37. a. The initial horizontal velocity is A (It's the only horizontal vector). b. The initial vertical velocity could be B (if projected down) or C (if projected upward). c. None of these; there is no horizontal acceleration. d. The vertical acceleration is B; it is always downwards. e. The net force on a projectile is B (there is only one force - gravity; and it is downwards).

  38. 2. Supposing a snowmobile is equipped with a flare launcher which is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)? • a. in front of the snowmobile • b. behind the snowmobile • c. in the snowmobile

  39. The answer is C. The flare will land in the snowmobile. The horizontal motion of the falling flare remains constant, and as such, the flare will always be positioned directly above the snowmobile. The force of gravity causes the flare to slow down and then return to the ground; yet it does not affect the horizontal motion of the flare.

  40. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane? • a. below the plane and behind it. • b. directly below the plane • c. below the plane and ahead of it

  41. The answer is B. The flare will land directly below the plane. The horizontal motion of the falling flare remains constant, and as such, the flare will always be positioned directly below the plane. The force of gravity causes the flare to fall but does not affect its horizontal motion.

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