A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

idku

  • one year ago

Ok, can someone show me ....

  • This Question is Closed
  1. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1441289020025:dw|

  2. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I found: Horizontal component of the velocity: \(\LARGE V_x=11\cos(44)\approx 7.9\) Vertical component of the velocity: \(\LARGE V_x=11\sin(44)\approx 7.6\) ---------------------------------------------------- How do I find the maximum height that the bsll goes above the ground?

  3. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    bsll is supposed to say ball.

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    What's the question ?

  5. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How do I find the maximum height that the bAll goes above the ground?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you know your equations of motion?

  7. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Which one?

  8. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I might as well just say, no I don't....

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    i think we need to know the height of those ppl

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    at least we need to know the height of the person throwing the ball

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is projectile motion correct, so it's better if you label your vector quantities more precisely, so we're dealing with the y - direction here\[\vec a = \frac{ v_{yf}-v_{yi} }{ t }\] where \[v_{yf}=0\] is there more information to this, all you gave is a diagram?

  12. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    f is final, i is initial. just clarifying....

  13. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I deleted the question I am so bump ---------------------------- Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

  14. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait, a is acceleration, but for maximum height, why do we need it? (btw, sorry again and again for deleted the question.)

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height

  16. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[v_y = u_y - gt\] set that equal to 0 and find the time at which that happens

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a = g in this case

  18. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what is u_y ?

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(a = -g\) because we take upwards as increasing position

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(u_y\) = initial vertical component of velocity

  21. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You can label it what ever, long as it makes sense to you and your teacher

  23. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    or, idk maybe I am confusing labels. I suck at this:)

  24. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    thats ridankulous normally we use \(v\) and \(u\) for velocity \(x\) and \(y\) for positions

  25. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah it is I that mixed something up (as always)

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we never used u, always v_0 xD

  27. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh yes V_(0y), that!

  28. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    v = u + at is how i still remember my kinematics eqns from highschool

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's much easier to

  30. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    it doesn't matter but we all need to agree on same thing..

  31. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}+9.81t\) and what did you say set equal to 0? like this, or am I wrong again?

  32. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    looks good and do you know why you're setting that equal to 0

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    no wait, you flipped the sign

  34. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}\color{red}{-}9.81t\) and what did you say set equal to 0? like this, or am I wrong again?

  35. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)

  36. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, you said -g, so I thought... ok will take a note

  37. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE 0=V_{0y}+9.81t\)

  38. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And I am setting it to 0, because V_y is the derivative of D_y (where D is displacement - in ase of y, the height)

  39. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So we are already having the derivative of the function that we want to maximize. Right?

  40. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, -9.81 ... my bad.

  41. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE 0=V_{0y}-9.81t\)

  42. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    remember this : \(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.

  43. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wait, so no minus; plus?

  44. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(a\) is acceleration, in our case \(a = -g = -9.81\)

  45. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\Large 0=V_{0y}\color{red}{\bf-}9.81t\)

  46. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so g is pos, -g is neg. at least I know that. I am a pes(t) - imist.

  47. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(a\) is negative because \(g\) is pointing downward but we are taking "upwards" as increasing position

  48. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok, and then I just find cthe maximum of our unknown displacement of y, based on the given velocity component (which is the derivative of the displacement)

  49. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    right, one thing at a time first find the time at which vertical component of velocity becomes 0

  50. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    t=0?

  51. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    at t=0 ... V_(0y) is 0.

  52. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    really ? at \(t=0\) the ball surely has nonzero vertical component of velocity : |dw:1441290728442:dw|

  53. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait, what is my equation for this? (to answer your question about the t-value at which the vertical velocity is equivalent to zero.)

  54. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Easy, as you said, vertical component of velocity of ball is modeled by the function : \[V_y=V_{0y}-9.81t \] you're "given" \(V_{0y} = 11\sin(44)\), therefore the function becomes \[V_y=11\sin(44)-9.81t \]

  55. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, can I try?

  56. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    let me finish typing first

  57. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure go ahead

  58. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    il let u work it but im not finished yet and i think u need to get some clarifications :)

  59. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you know that \(V_y\) is \(0\) when the ball reaches maximum height so use that fact to find the "time" at which the ball reaches maximum height

  60. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    once you have the "time" at which ball reaches the maximum height, you can use that to find the required maximum height

  61. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is a nice way to work with vertical trajectory you could also still do it the way I was showing above, the main thing is knowing the final velocity of the y component = 0 at maximum height

  62. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    that is all i wanted to say go ahead, set \(V_y\) equal to \(0\) and solve \(t\)

  63. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE V_y=11\sin(44)-9.81t \) \(\LARGE 0=11\sin(44)-9.81t \) \(\LARGE 11\sin(44)=9.81t \) \(\LARGE t=\frac{9.81}{11\sin(44)} \)

  64. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hope at least that I have managed:)

  65. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm

  66. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh no

  67. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well that to the power of -1

  68. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It should be \[t = \frac{ 11\sin(44) }{ 9.81 }\]

  69. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE t=\frac{11\sin(44)}{9.81} \) I am even dumber than what I thought

  70. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, I am dividing by 9.81 don't know what's up with my Brain

  71. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78 \)

  72. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You could actually go ahead and use \[y = v_{0y} t\] now

  73. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That will give you your maximum height

  74. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96\)

  75. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.

  76. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That is the max height....

  77. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Let me make an attempt to recap what I have done for maximum height ((My suffering is not equivalent of any sin (of \(\theta\)), I hate this!)) ----------------------------------------------------- RECAP, MAX HEIGHT \(\LARGE V_y=V_{0y}-gt \) \(\LARGE 0=11\sin(44)-9.81t \) \(\LARGE t =0.78\) so max height occrs at t=0, and now we plug that into vertical displacement \(\LARGE y=V_{0t}\times t \) \(\LARGE y=11\sin(44)\times 0.78=5.96 \)

  78. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    t=0?

  79. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, what do you mean?

  80. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why did you write max height is at t = 0

  81. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, occurs at t=0.78

  82. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Don't round until the final answer

  83. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So use the full time rather than 0.78

  84. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\LARGE y=\sin^2(44)/9.81\) ?

  85. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I forgot 11

  86. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    0.78 was really 11sin(44)/9.81 So y is y=11sin(44)×11sin(44)÷9.81

  87. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well, if that is right then it is still 5.95

  88. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\Large y = v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = 11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\) http://www.wolframalpha.com/input/?i=11sin%2844%29%C3%97%2811sin%2844%29%2F9.81%29-%281%2F2%29%289.81%29%2811sin%2844%29%2F9.81%29%C2%B2 I get 2.97597 ----> 2.98

  89. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And that is not correct either -:(

  90. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hold on, let just use \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) }\] we also have to remember the 1.5m forgot about that

  91. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't think that there is another explanation for an incorrect output from the formula, other than that I failed to find all the values correctly. I don't think formulas are faulty. (I start to despair from thinking that I can think ) HIHPHh9hduhoduhdue89wv9vi

  92. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I have to go, I will come back about 2 h later-;(

  93. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, this should work, the problem was we did not take account of the question, you must read it carefully, and not make assumptions. Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) } = \frac{ [11\sin(44)]^2 }{ 2(9.81)} = 2.975...\] but we must add the 1.5m as well to this, so we get 2.975...+1.5 = 4.4759m = 4.5m

  94. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think you should use this though as we found time for this precisely \[h = \frac{ 1 }{ 2 }g t^2 = \frac{ 1 }{ 2 }(9.81)(0.78)^2 = 2.965....+1.5 = 4.47 = 4.5m\] so it works!

  95. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    below equation for vertical position perfectly fine \(\Large y = \color{red}{y_0}+v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = \color{red}{y_0}+11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\)

  96. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When you put your final answer, make sure it's 2 sig digs!

  97. idku
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, if it was something on a same level of thinking then I would have probably figured. Yes the 1.5 meters of the ground, because my origin is (0, 1.5) Iambatman tnx! I am checking it and it is.....

  98. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.