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Omniscience Group Title

Projectile Motion: Two objects are projected with the same speed simulataneously. The object A is projected horizontally from the height H and B vertically from the ground at mid range of A. The objects later collide at some height above the ground. (will draw diagramn below)

  • one year ago
  • one year ago

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  1. Omniscience Group Title
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    |dw:1343399513614:dw|

    • one year ago
  2. Omniscience Group Title
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    Show that \(R = 2H\)

    • one year ago
  3. Omniscience Group Title
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    well how do i start? anyone able to give me a hint?

    • one year ago
  4. agentx5 Group Title
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    Where to start? With a FBD of course! (free body diagram) ^_^ That's the easy step Show the forces as vectors acting on each object

    • one year ago
  5. Omniscience Group Title
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    i dont know vectors..

    • one year ago
  6. agentx5 Group Title
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    The goal of that first step is so you can see what is acting on a frame of reference What forces are acting on A? What forces are acting on B? Can you make two parametric equations for these? (you can) :-)

    • one year ago
  7. agentx5 Group Title
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    |dw:1343400018031:dw|

    • one year ago
  8. Omniscience Group Title
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    I am only able to form parametric equation for \(A\) \[y=-\frac{1}{2} g*\left(\frac{x}{u}\right)^{2}+H \]

    • one year ago
  9. agentx5 Group Title
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    The "down" is gravity acting on each object's mass, in other word the weight. The up arrow and right arrow are the initial directions of motion

    • one year ago
  10. agentx5 Group Title
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    Consider the horizontal and vertical motions of each: \(\sum \text{Forces}_y = \ ?\) , \(\sum \text{Forces}_x = \ ?\) Which wins out? Gravity right? What's the motion of object A for example? Upside down parabola right? Why?

    • one year ago
  11. agentx5 Group Title
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    This is actually something you can solve in your head by just thinking about it visually. The motion of object A is going to be (ignoring air friction, which if you wanted to be sneaky you could use to disprove the question how it was asked as false btw), moving down at an accelerating rate, t\(^2\); whereas the horizontal motion is strictly a linear rate constant, t. What you're finding here is nothing more than the intersection of two parametric curves. The parameter here is time, and there's an x & y equation for each object. The part that makes this easier is that equation for the x (horizontal) direction for object B is literally \(F_{2,x}(t) : x_2 = 0\). Am I making sense?

    • one year ago
  12. Omniscience Group Title
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    well what do i do next? I am unable to find the trajectory equation thing for \(B\)

    • one year ago
  13. experimentX Group Title
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    what about the velocity?

    • one year ago
  14. Omniscience Group Title
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    i am able to find the acceleration, velocity for \(A\) but not \(B\)

    • one year ago
  15. experimentX Group Title
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    no ... i'm asking about the initial velocity of both projectiles

    • one year ago
  16. Omniscience Group Title
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    i dont know; 0?

    • one year ago
  17. experimentX Group Title
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    what's this R anyway?

    • one year ago
  18. Omniscience Group Title
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    R is the range; the longest horizontal distance travelled by the object

    • one year ago
  19. Omniscience Group Title
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    |dw:1343402352605:dw| abit clearer; sorry for the horrible drawing

    • one year ago
  20. experimentX Group Title
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    trivial case v=0, R=0, H=0 ... what' the initial velocity of both projectiles .. are they same?

    • one year ago
  21. Omniscience Group Title
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    well the velocity i got for A for the x-motion is \(u\) and the y-motion is \(-gt\)

    • one year ago
  22. experimentX Group Title
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    what about the B?

    • one year ago
  23. Omniscience Group Title
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    i dont know how to do B

    • one year ago
  24. Omniscience Group Title
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    because its projected vertically so im lost

    • one year ago
  25. experimentX Group Title
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    i mean what is the initial velocity of B ... going up?

    • one year ago
  26. Omniscience Group Title
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    i dont know..i guess its going up..

    • one year ago
  27. experimentX Group Title
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    Eqn for 1 x = ut y = H -1/2 gt^2 Eqn for 2 y = 1/2 gt^2 x = R/2 solve those two R/2 = ut t = R/2u y = H - 1/2 g (R/2u)^2 = H - gR^2/8u^2 --- 1 y = gR^2/8u^2 --- 2 H - gR^2/8u^2 = gR^2/8u^2 H = 1/4 gR^2/u^2

    • one year ago
  28. Omniscience Group Title
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    how did you get y = (1/2)gt^2 for the second equation?

    • one year ago
  29. experimentX Group Title
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    v^2 = y^2 + 2as => H = u^2/2g

    • one year ago
  30. Omniscience Group Title
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    okay; thanks!

    • one year ago
  31. experimentX Group Title
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    H = 1/2R^2/H => 2H = R always make sure you ask question correctly. the equation of motion of projectile is given by \[ x = u_xt \\ y = v_y + 1/2 gt^2 \] in your case horizontal component is absent in one while vertical is absent in another.

    • one year ago
  32. experimentX Group Title
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    and the horizontal v of one was u while vertical velocity of another was u ... more accurately, the parametric equation of motion is given by \[ x = x_0 + ... \\ y = y_0 + ... \] if they collide means they you have to equate x's and y's of 1 and 2 ... you should be able to solve them

    • one year ago
  33. Omniscience Group Title
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    thank you!

    • one year ago
  34. experimentX Group Title
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    yw

    • one year ago
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