A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

idku

  • one year ago

Math - Based - Physics question.

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

    |dw:1441307711913:dw||dw:1441308026115:dw|

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

    `NOTE: in the corner of the picture 2, all it says is "1.5 m".`

  3. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well you now know that its horizontal velocity is 9m/s and that it travels a vertical distance of 5.5m before it stops rising that second piece of info allows you to calculate its vertical velocity as it leaves her hand use the equation of motion \(v^2 = u^2 + 2ax\) v = 0, a = -g, x = 5.5m, so get u

  4. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you follow? or questions?

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

    First of, how do I know that it travles 5.5 meters before it stops rising?

  6. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the question says it rises to 7m above ground, it start from 1.5m above ground, 7 - 1.5 = 5.5m

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

    Oh, ok, 5.5m is our vertical displacement. Good. And what does "u" stand for?

  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    u = initial velocity v = final velocity yet another notation but it's simple and really easy on the eye eg because no subscripts

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

    Oh, good. And v=0 because that is the point where it changes from positive (the height of the ball increasing) to negative (the height of the ball decreasing). Am I right?

  10. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes

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

    \(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(-9.81)(5.5)\) like this?

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

    or, am I plugging +9.81 ?

  13. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you got it right!

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

    the first time?

  15. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this is right \[\large 0^2=u^2+2(-9.81)(5.5)\]

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

    \(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(-9.81)(5.5)\) \(\large 0^2=u^2-107.91\) \(\large u=\sqrt{107.91}\)

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

    yes, if I plugged in +9.81 I would have got imaginary velocity.

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

    u=10.39 m/s (this is initial velocy)

  19. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    *yes, if I plugged in +9.81 I would have got imaginary velocity.* nice! i'm not doing the sums so i trust you are happy with your calculation?

  20. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    OK, really important now. right now, we are working in the vertical direction! yes?!

  21. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so "u=10.39 m/s (this is initial VERTICAL velocy)" [assuming your calculator is working!]

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

    I used wolfram, so should work, but the practice thingy doesn't accept 10.39

  23. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we already know the HORIZONTAL velocity, which is constant yes?

  24. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we have to combine vertical and horizontal using pythagoreas

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

    I am not given the angle, how would I tell horizontal velocty/

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

    ?

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

    Sorry.

  28. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i told you it in my first post go take a look and see where it came from

  29. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    in case you're not getting this i will repeat it:p we already know the HORIZONTAL velocity, which is constant yes? i told you it in my first post go take a look and see where it came from bandwidth is an issue for me :-(

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

    I got disconnected due to lightning. My apologies, I am here....

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

    yes, I see so I got 13.75 m/s http://www.wolframalpha.com/input/?i=%E2%88%9A%289%C2%B2%2B10.39%C2%B2%29

  32. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no worries! hope everyone's OK.

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

    No one is outside:) My garage is inside the house, so it's alright.

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

    I wish I had you as my physics teacher, you made everything as easy as even a non-math person such as myself got it!

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

    There is just one more question (for this part at least), if you don't mind: `How high above the ground will the ball be when it gets to Julie?`

  36. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    excellent, one thing you **must** remember we do the horizontal axis and vertical differently no acceleration in horizontal, g acts in the vertical then combine through pythagoreas that applies throughout mechanics and kinematics...

  37. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well you know how long it will take to get to Julie, right

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

    ok, i will take a note of pythagorean as regards to velocities as well as distances....

  39. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    brill! you know ho far apart they are? you know the horizontal velocity? and so?

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

    Horizontal velocity is 9m/s. And the distance between them would be?

  41. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    didn't you work that out in an earlier part of the question?

  42. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    use that number

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

    I did the maximum height for [the part in] picture 1, and I did horizontal and vertical displacements for that (and this) part(s).

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

    wait, is it 11cos(44) ?

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

    but that is the velocity of the first through, not the distance. So that is not it.

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

    9(the horiz. velocity of the first through))

  47. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    let me work it out first, it might jog your memory just wait

  48. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    OK from the very first question, we know that it was thrown at 11m/s at 44deg we can use equation ofmotion to get flight time \[x = ut + \frac{1}{2}a t^2 = t(u + \frac{1}{2}at)\] follow so far? we are working in the vertical direction now.

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

    oh, so the displacement-for-y formula and displacement-for-x formula are all same, besides that every y is now x instead?

  50. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    always in these x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time always here we are tracking the ball from the first girl to the second, and we are looking at motion in the vertical direction, where gravity plays a role

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

    wait, you said: \(x = ut + \frac{1}{2}at^2\) so x in this case is the vertical distance I am solving for. `a = -g = -9.81m/s` u = 9m/s ` t = ?` how do I figure out time?

  52. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we need to back to the first question to get the distance between the girls, because we know that one threw it and one caught it. so we know it started t 1.5m above ground, and ended there when caught. we also know it was thrown at 11m/s at 44drg to horizontal.

  53. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and to do that we need first to calculate how long the ball spent in the air travelling between the girls there's a lot in there so take your time

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

    While the distance is same, the times are different at different speeds. So we can't use the first question to calculate the time of second through. (That would have been rediculous if I have done so).

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

    But, distance I can use that same displacement formula, applying everything I know from picture 1?

  56. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no we need to know how far the girls are apart we cannot do that from the information in the second part of the question we need to go right back to the first part to do that that is why i suspect that you have already calculated that distance

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

    distance = 12.32

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

    meters*

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

    \(12.32 = 9t + \frac{1}{2}(-9.81)t^2\) ?

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

    and olve for time?

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

    solve

  62. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+9.81 correct

  63. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    OK so back to this last bit and we need to recap, i think. we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s yes now how long will it take to reach the first girl? in the horizontal direction....

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

    Oh, I am setting incorrect equation, because we are working the time based on the information we have from the first question, so u=11

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

    and I said u=9

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

    \(12.32 = 11t + \frac{1}{2}(-9.81)t^2\) But I still get imaginary solutions

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

    with +9.81 I get real solutions:)

  68. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    just stop for a minute, i think we are crossing posts:p we know the distance between them is 12.327m we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s agreed?

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

    yes, we had all of this either as given or found that:)

  70. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so now we need to know how long that will take, for a ball with horizontal velocity = 9m/s to travel the distance between the girls, which is 12.327m

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

    \(12.327= 9t + \frac{1}{2}(-9.81)t^2\)

  72. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    there is no gravity in the horizontal direction:p

  73. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    just constant velocity in horizontal direction

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

    what is my formula for x then?

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

    which formula do I use, I mean

  76. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oooHHHHH!!!! think!

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

    \(x = ut + \frac{1}{2}t^2\) ?

  78. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the first one you learn!

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

    Distance = Velocity × Time ?

  80. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[x = ut + \frac{1}{2}(0)t^2 = ut\] or t = distance / velocity !!!!

  81. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah!!

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

    oh, for horizontal veloccity the acceleration is 0. distance = (initial velocity) × (time) ---> time = distance ÷ (initial velocity) time = 12.327m ÷ 9m/s (when dividing by m/s, you multiply times s/m, so meters cancel we get: (12.327/9)s )

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

    s is seconds.

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

    for whatever that quotient is

  85. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes, 1.37s http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+%289.81+*+9%29

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

    ≈ 1.37

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

    yes:)

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

    Now we need to apply t=1.37 (time length interval of the ball's second through) to find the height of the ball when it gets to julie - that is when it travels 12.327 meters.

  89. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    one final recap before we finish we know the distance between them is 12.327m [but we don't need that info anymore] we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s we know that the journey in th e**horizontal** direction too 1.37s. we need to know what happens in the vertical direction at t = 1.37. at what height is the ball? agreed?

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

    yes, I undertand that! You are a wonderful person!

  91. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so we need an equation of motion with acceleration in it because of gravity \[x = ut + \frac{1}{2}at^2\]

  92. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we want to know x. x is the displacement of the ball in the vertical direction a = -9.91 m/s/s t = 1.37 s u = 10.39 m/s

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

    x is the variable of interest. `initial velocity -- u=10.39m/s (we are interested in vertical velocity)` time -- t=1.37s (time is same for Vert. and Horiz. displacement of the ball) `acceleration [due to gravity] -- a=-(g)=-(9.81)`m/s² \(x = 10.39m/s \times 1.37s + \frac{1}{2}(-9.81m/s^2)(1.37s)^2\)

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

    so in the first and second part we have meters left only. Just making sure t makes sense the way I do it, so that if I encounter unit confusion I know I have done something wrong. \(x = 10.39 \times 1.37 + \frac{1}{2}(-9.81)(1.37)^2\)

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

    and whatever that is equal to....

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

    5.028m

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

    Verif. http://www.wolframalpha.com/input/?i=12.327%3D9t%2B%281%2F2%29%289.81%29t%C2%B2

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

    +1.5m

  99. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes so she throws it 5.03m above the first girls hand. add 1.5m onto that if you want distance above ground

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

    I get 5.028+1.5=6.528

  101. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh! good you got it yourself:p

  102. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well, that's the right "way" to do it, any mistakes are calculations.....we now know "how" to do it:p

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

    so the main things I need are: \(\large v^2=u^2+2ax\) \(\large x=ut+\dfrac{1}{2}at^2\)

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

    and of course I know how to find horizontal and vertical velocity components based on a given angle and initial velocity....

  105. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    let me type them out, there are 3 \[v = u + at\] \[x = ut + \frac{1}{2}at^2\] \[v^2 = u^2 + 2ax\] trick is to pick the right one to use as they all look at different things. eg the last one doesn't have t in it. but it is incredibly useful if you know of it.

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

    this is probably the most valuable post in my life. TY so much. No excuse me I got to go:) tnx for formula recap!

  107. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes - horizotal vs vertical there's an awful awful lot of info in there, you have done well! cheers!!

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

    \(v = u + at\) \(v_x~~ =~~u~ \cos(\theta)\) \(v_y ~~=~~ u~\sin(\theta)\) \(d = ut + \frac{1}{2}at^2\) \(v^2 = u^2 + 2ax\) x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time Ok, when I have to time to look more into it, i will. Thank you very much. I actually get this stuff now.

  109. 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.