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YumYum247

  • one year ago

A soccerball was kicked at an angle of 28 deg to the ground and it landed 39.8m away. a) what was the velocity of the ball when kicked?

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  1. YumYum247
    • one year ago
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    @Astrophysics just give me a quick hint and leave the rest to me.....

  2. Astrophysics
    • one year ago
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    VECTORS PROJECTILE MOTION YAYAYAYA DIAGRAMS

  3. YumYum247
    • one year ago
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    |dw:1439958921010:dw|

  4. Astrophysics
    • one year ago
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    |dw:1439959033896:dw| I wanted to make an arc

  5. YumYum247
    • one year ago
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    just tell me what's the first thing i need to look for....that's all

  6. Astrophysics
    • one year ago
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    |dw:1439959053044:dw| now lets make it nice and clean so we know exactly what's going on, we need to find v0 here which is velocity of the ball when it was kicked.

  7. Astrophysics
    • one year ago
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    You need to find your velocities at x and y direction, play around with it and apply kinematic formulas

  8. YumYum247
    • one year ago
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    yes but for that i need to find the final velocity :(

  9. YumYum247
    • one year ago
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    |dw:1439959268885:dw|

  10. Astrophysics
    • one year ago
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    We don't actually need it, I have an idea!!!!

  11. Astrophysics
    • one year ago
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    Want to hear it

  12. YumYum247
    • one year ago
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    yuss!!!!

  13. YumYum247
    • one year ago
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    Vertical: Horizontal: a = -9.8m/sec2 dh = 39.8m Dh = 0m/s Vh = ? Vi = ? t = ? Vf = ? t = ?

  14. Astrophysics
    • one year ago
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    Ok I was actually thinking it gave us the maximum height

  15. YumYum247
    • one year ago
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    nope that's all we got???

  16. Astrophysics
    • one year ago
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    Good lets put everything in x and y direction...ok lets see and thing and have fun!

  17. YumYum247
    • one year ago
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    sorry i forgot to add the theta in there... Vertical: Horizontal: a = -9.8m/sec2 dh = 39.8m Dh = 0m/s Vh = ? Vi = ? t = ? Vf = ? t = ? Angle = 28deg

  18. Astrophysics
    • one year ago
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    x - direction x = 39.8 m \[\large v_{x_0} = v_0 \cos (28)\] y-direction a = -g \[\large v_{y_0} = v_0 \sin(28)\] so this is what we're given hahahaaa

  19. YumYum247
    • one year ago
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    yes....

  20. Astrophysics
    • one year ago
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    Ok ok we can actually use the range formula

  21. Astrophysics
    • one year ago
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    and find initial velocity instantly XD

  22. Astrophysics
    • one year ago
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    \[R = \frac{ v_0^2\sin(2 \theta) }{ g }\]

  23. YumYum247
    • one year ago
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    hey about that..... what does it mean by Vi here, is it the horizontal velo or the vertical???

  24. Astrophysics
    • one year ago
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    Oh no that's the resultant velocity |dw:1439960011427:dw| what we're looking for

  25. Astrophysics
    • one year ago
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    That can be split into components as shown to figure it out

  26. YumYum247
    • one year ago
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    yah cuz that formula says dh = Vi^2 X Sin2theta/a so that's why i was wondering what does that initial velocity means here....

  27. YumYum247
    • one year ago
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    horizontal or vertical...cuz if it's the resultant then shouldn't it be Vf^2????

  28. Astrophysics
    • one year ago
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    I don't follow, that's the initial velocity, we are looking for the velocity that it was kicked at not at the velocity it landed at

  29. YumYum247
    • one year ago
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    |dw:1439960370291:dw|

  30. Astrophysics
    • one year ago
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    Oh I see what you're trying to do, no that's not right, because we're splitting the initial velocity into components

  31. Astrophysics
    • one year ago
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    |dw:1439960455619:dw|

  32. Astrophysics
    • one year ago
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    Those are the components of the initial velocity, does that make sense?

  33. Astrophysics
    • one year ago
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    you can label it as i for initial it's the same thing :P

  34. YumYum247
    • one year ago
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    those nots confuse me ......i only understand Vf, Vi and Vh

  35. Astrophysics
    • one year ago
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    Ok sorry ill fix it

  36. YumYum247
    • one year ago
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    Thanks :")

  37. Astrophysics
    • one year ago
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    |dw:1439960585668:dw| same thing

  38. YumYum247
    • one year ago
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    oh now i see...... :D

  39. Astrophysics
    • one year ago
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    Yeah they are just components of the initial velocity :-)

  40. Astrophysics
    • one year ago
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    \[R = \frac{ v_i^2\sin(2 \theta) }{ g }\]

  41. YumYum247
    • one year ago
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    ok so where are we going with that formula, are we solving for the Vi there???

  42. Astrophysics
    • one year ago
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    Yup!

  43. YumYum247
    • one year ago
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    ok so it'll be like this....

  44. YumYum247
    • one year ago
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    |dw:1439960730234:dw|

  45. YumYum247
    • one year ago
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    do we take the square root??? O_o

  46. Astrophysics
    • one year ago
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    Mhm hold on a sec, lets just do the algebra first to avoid confusion the range here is our x or distance which is 39.8 m. It's always best to do algebra first then plug things in especially when you get into deeper physics, it will benefit you. \[x= \frac{ v_i^2 \sin(2 \theta) }{ g }\] now solve for the initial velocity

  47. Astrophysics
    • one year ago
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    Don't plug anything in just solve for initial velocity

  48. Astrophysics
    • one year ago
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    variable

  49. YumYum247
    • one year ago
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    ok so rearrange the variables first....

  50. Astrophysics
    • one year ago
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    Yup

  51. YumYum247
    • one year ago
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    and then plug i nthe numbers....???

  52. Astrophysics
    • one year ago
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    Yeah

  53. YumYum247
    • one year ago
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    aight thanks man, i really appreciate your help..... Stay cute and blessed!!!! :")

  54. Astrophysics
    • one year ago
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    Np, lets just finish it off :P, so what do you get when you solve for initial velocity when you rearrange don't worry about making mistakes, that's how we learn!

  55. YumYum247
    • one year ago
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    hold up.....

  56. Astrophysics
    • one year ago
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    Take your time

  57. YumYum247
    • one year ago
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    x = Vi^2 X Sin2theta/a x.a = Vi^2 X Sin2theta x.2/sin2Theta = Vi^2

  58. Astrophysics
    • one year ago
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    Perfect now just square root it and we're good!

  59. YumYum247
    • one year ago
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    |dw:1439961336297:dw|

  60. Astrophysics
    • one year ago
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    I'm assuming that 2 in the numerator was meant to be an a haha

  61. YumYum247
    • one year ago
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    ok man thanks alot i really appreciate your effort.... stay blessed!! :)

  62. Astrophysics
    • one year ago
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    |dw:1439961409567:dw| whats that

  63. YumYum247
    • one year ago
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    x

  64. YumYum247
    • one year ago
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    or R

  65. YumYum247
    • one year ago
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    39.8m

  66. Astrophysics
    • one year ago
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    Hmm...our equation was \[x= \frac{ v_i^2 \sin(2 \theta) }{ g } \implies v_i = \sqrt{\frac{ gx }{ \sin(2 \theta) }}\]

  67. Astrophysics
    • one year ago
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    I think you may have added an extra x by mistake

  68. YumYum247
    • one year ago
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    yah!!! thats it....!!!

  69. YumYum247
    • one year ago
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    no that's just my hand writing :"D

  70. YumYum247
    • one year ago
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    ok thanks and good night, sleep tight and don't let the bugs bite :"D

  71. Astrophysics
    • one year ago
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    Take care :)

  72. YumYum247
    • one year ago
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    too 2 :)

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