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Omniscience

  • 2 years ago

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)

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  1. Omniscience
    • 2 years ago
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    |dw:1343399513614:dw|

  2. Omniscience
    • 2 years ago
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    Show that \(R = 2H\)

  3. Omniscience
    • 2 years ago
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    well how do i start? anyone able to give me a hint?

  4. agentx5
    • 2 years ago
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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

  5. Omniscience
    • 2 years ago
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    i dont know vectors..

  6. agentx5
    • 2 years ago
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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) :-)

  7. agentx5
    • 2 years ago
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    |dw:1343400018031:dw|

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

  9. agentx5
    • 2 years ago
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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

  10. agentx5
    • 2 years ago
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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?

  11. agentx5
    • 2 years ago
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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?

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

  13. experimentX
    • 2 years ago
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    what about the velocity?

  14. Omniscience
    • 2 years ago
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    i am able to find the acceleration, velocity for \(A\) but not \(B\)

  15. experimentX
    • 2 years ago
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    no ... i'm asking about the initial velocity of both projectiles

  16. Omniscience
    • 2 years ago
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    i dont know; 0?

  17. experimentX
    • 2 years ago
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    what's this R anyway?

  18. Omniscience
    • 2 years ago
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    R is the range; the longest horizontal distance travelled by the object

  19. Omniscience
    • 2 years ago
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    |dw:1343402352605:dw| abit clearer; sorry for the horrible drawing

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

  21. Omniscience
    • 2 years ago
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    well the velocity i got for A for the x-motion is \(u\) and the y-motion is \(-gt\)

  22. experimentX
    • 2 years ago
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    what about the B?

  23. Omniscience
    • 2 years ago
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    i dont know how to do B

  24. Omniscience
    • 2 years ago
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    because its projected vertically so im lost

  25. experimentX
    • 2 years ago
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    i mean what is the initial velocity of B ... going up?

  26. Omniscience
    • 2 years ago
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    i dont know..i guess its going up..

  27. experimentX
    • 2 years ago
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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

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

  29. experimentX
    • 2 years ago
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    v^2 = y^2 + 2as => H = u^2/2g

  30. Omniscience
    • 2 years ago
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    okay; thanks!

  31. experimentX
    • 2 years ago
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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.

  32. experimentX
    • 2 years ago
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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

  33. Omniscience
    • 2 years ago
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    thank you!

  34. experimentX
    • 2 years ago
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    yw

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