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anonymous

  • one year ago

A SWAT team who is 10 meters away from the bus launches a tear gas that hits its wrecked window at point 2.5 meters above the ground. IF the tear gas is projected at an angle 38 degrees from the horizontal, determine its total time of flight. What is the initial velocity of the tear gas? Sketch the problem.

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  1. anonymous
    • one year ago
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    Going to try the problem until someone helps me.

  2. anonymous
    • one year ago
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    |dw:1439370371073:dw|

  3. anonymous
    • one year ago
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    is my sketch right?

  4. Jack1
    • one year ago
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    who cares if it's right... that sketch is epic art dude! props~!!!

  5. anonymous
    • one year ago
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    especially the stickman? XD

  6. Jack1
    • one year ago
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    totes ;)

  7. Jack1
    • one year ago
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    10m away = yep 2.5 high window = yep i guess we're assuming they launch the gas thing from ground level?? 38 degrees from horizontal... yep drawing looks great now, the important thing: can u solve it?

  8. anonymous
    • one year ago
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    First thing I thought about this problem

  9. anonymous
    • one year ago
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    component method. The vy= vosintheta

  10. anonymous
    • one year ago
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    and vx=vocostheta

  11. anonymous
    • one year ago
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    I assume I'll get the magnitude? Then I'll use the formula Vy = Voy - gt ?

  12. anonymous
    • one year ago
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    Ohhh wait that's wrong

  13. anonymous
    • one year ago
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    First thing is I need to know the y-component

  14. anonymous
    • one year ago
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    |dw:1439371607809:dw|

  15. Jack1
    • one year ago
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    |dw:1439371463873:dw|

  16. anonymous
    • one year ago
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    Hmmm.. Do you I think I need to get the max height? (if it is possible)

  17. Jack1
    • one year ago
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    i guess final height = 2.5m initial height = 0 m max height = ?

  18. anonymous
    • one year ago
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  19. Jack1
    • one year ago
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    distance horizontal = 10m... initial speed = ?? acceleration vertical = -9.8 m/s i guess that's all we have to go on... yeah?

  20. anonymous
    • one year ago
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    Yeaaap soooo

  21. anonymous
    • one year ago
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    I posted a formula but I don't know if we can use it due to Vo is unknown too :(

  22. Jack1
    • one year ago
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    so ur going to have to rephrase the time in terms of the horizontal distance, then plug that into ur max height calc... sounds like a mess of work...

  23. anonymous
    • one year ago
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    \[y=Vot-1/2(g)(t)^2\]

  24. anonymous
    • one year ago
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    Okayy you have other ways I'm quite confuse

  25. Jack1
    • one year ago
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    give me 15... just have to dinner, then we'll solve

  26. anonymous
    • one year ago
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    Will do. :D

  27. anonymous
    • one year ago
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    Another viewer welcome!!

  28. Jack1
    • one year ago
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    hey dude, back so: known factors: initial height = 0 final height = 2.5m max height = ? initial distance = 0 final distance = 10m acceleration: ay = -9.8 m/s^2 ax = 0 Initial Speeds: Voy = Vo x sin 38 Vox = Vo x cos 38 Distance equations: sy = Voy * t + 1/2 ay t^2 sx = Vox * t + 1/2 ax t^2

  29. anonymous
    • one year ago
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    okay game

  30. Jack1
    • one year ago
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    sx = Vox * t + 1/2 ax t^2 and ax = 0 sx = Vox * t + 0 10 = Vox * t t = 10/Vox and Vox = Vo * cos 38 so t = 10/Vox t = 10/(Vo * cos 38) sy = Voy * t + 1/2 ay t^2 2.5 = Voy * t + 1/2 ay t^2 2.5 = Voy * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2 and Voy = Vo x sin 38 so 2.5 = Voy * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2 2.5 = (Vo x sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

  31. Jack1
    • one year ago
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    what a freakin mouthful that is! ;D so... can you solve that in terms of Vo from here?

  32. Jack1
    • one year ago
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    ooo... also: does that make sense? any issues with those kinematic equations dude?

  33. anonymous
    • one year ago
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    Nah nah I'm good will read your text first gimme a sec

  34. anonymous
    • one year ago
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    not literally a sec but maybe longer lol

  35. Jack1
    • one year ago
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    |dw:1439376204600:dw|

  36. anonymous
    • one year ago
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    wait still reading and understanding

  37. anonymous
    • one year ago
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    Question about the distance equation

  38. anonymous
    • one year ago
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    sy = Voy * t + 1/2 ay t^2 isn't it sy= Voy*t - 1/2g(t)^2 ??

  39. anonymous
    • one year ago
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    or maybe u just translate it to other terms and it is just the same

  40. anonymous
    • one year ago
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    nvm the question I got it

  41. Jack1
    • one year ago
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    yep, its the same, i just use a for acceleration rather than g for gravity

  42. anonymous
    • one year ago
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    in freefall ax is always 0 right?

  43. anonymous
    • one year ago
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    bc there's force upon it only the y-component

  44. anonymous
    • one year ago
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    no force*

  45. Jack1
    • one year ago
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    yep, we assume wind resistance = negligable, so no x component to acceleration

  46. anonymous
    • one year ago
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    Can you explain the logic behind putting t=10/voxcos38 in the sy=voyt+1/2ay(t)^2

  47. Jack1
    • one year ago
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    because we want an equation with only 1 unknown in it... so we put everything in terms of inital velocity (Vo - not component split) then we solve for Vo then we use that to solve for time

  48. anonymous
    • one year ago
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    2.5 = (Vo x sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

  49. anonymous
    • one year ago
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    isn't it that vo is also vox you just forgot to put x on it?

  50. anonymous
    • one year ago
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    okay we're finding vox after this

  51. anonymous
    • one year ago
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    no wait vox = 10 right?

  52. Jack1
    • one year ago
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    no, we're finding Vo Vox is the x component (horizontal component) of Vo we just want Vo

  53. anonymous
    • one year ago
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    lol jk

  54. anonymous
    • one year ago
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    sx = 10

  55. anonymous
    • one year ago
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    moving on I get it

  56. Jack1
    • one year ago
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    2.5 = (Vo * sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

  57. anonymous
    • one year ago
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    Okay moving on after I got Vo Ill use this?

  58. anonymous
    • one year ago
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  59. anonymous
    • one year ago
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    2(vo)sin(38)/g

  60. Jack1
    • one year ago
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    just solve Vo first then put that back into t = 10/(Vo * cos 38) to solve for t then... well ... done, yeah? ur time in flight equation assumes the start and finish height are the same... so it wont work for this question

  61. anonymous
    • one year ago
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    Oh yeah.

  62. anonymous
    • one year ago
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    I forgot about that

  63. anonymous
    • one year ago
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    The question is quite odd I thought we need to find the T first before Vo. But ohwell what works.. works

  64. Jack1
    • one year ago
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    focus dude, solve Vo

  65. anonymous
    • one year ago
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    Ohmy I thought we are done. Sorry!! Forgive me. Wait a sec I'll solve

  66. Jack1
    • one year ago
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    lol, sçool, jus wanna make sure we get the same answers for Vo and t, s'all ;)

  67. anonymous
    • one year ago
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    I got vo=12.1871

  68. Jack1
    • one year ago
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    perfect (+ or -... and minus is insane, so answer is positive 12.1871) now t?

  69. anonymous
    • one year ago
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    1.0412 well that was fast

  70. anonymous
    • one year ago
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    such a skilled swat team LOL

  71. Jack1
    • one year ago
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    for a grenade... psh, not that fast like olympic sprinters = 100m in sub ten seconds... so yeah... sprint speed

  72. anonymous
    • one year ago
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    well if you think of it under pressure and you're throwing it in a window! Such accuracy

  73. Jack1
    • one year ago
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    totes well, i'm out, catchya dude, u cool with all that?

  74. anonymous
    • one year ago
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    Of course thankyou!! U have yourself a new fan :D

  75. Jack1
    • one year ago
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    logend, no worries bro, slaters ;P

  76. Jack1
    • one year ago
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    *legend

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