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anonymous

  • one year ago

Physics stuff!! A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

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  1. anonymous
    • one year ago
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    @ganeshie8 @nincompoop @pooja195 @Preetha any ideas?

  2. Empty
    • one year ago
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    I believe there are two ways to do this 1) Calculate the trajectory 2) Calculate the energy change

  3. ganeshie8
    • one year ago
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    presuming the ball lands on the ground and not on the cliff... even though it was thrown vertically upwards

  4. anonymous
    • one year ago
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    this is a calc I problem by the way

  5. ganeshie8
    • one year ago
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    maybe the angle shouldn't matter here because the ball gains its speed by the time it reaches cliff. You may start with the given acceleration due to gravity : \[v(t)-v(0)=\int\limits_0^t a(t)\,dt \]

  6. anonymous
    • one year ago
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    okay s v(0) = 40 but how do we find v(t) ?

  7. Empty
    • one year ago
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    Continue further, \[v(t) = \frac{d x(t)}{dt}\]

  8. anonymous
    • one year ago
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    wait slow down , how did you find s(t) ?

  9. ganeshie8
    • one year ago
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    \[\begin{align}v(t)-v(0)&=\int\limits_0^t a(t)\,dt \\~\\ &=\int\limits_0^t -32\,dt \end{align}\] you're given that \(a(t)=-32\), see if you can integrate above now..

  10. anonymous
    • one year ago
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    It is position. 500ft cliff above the ground=+500 Thrown upward with speed of 40 feet per second-it will eventually fall down to the ground which is negative=40t acceleration is negative because velocity is negative -32 feet per second squared t is seconds so second squared=-32t^2

  11. anonymous
    • one year ago
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    v(t) - 40 = (-32t^2)/2 ?

  12. anonymous
    • one year ago
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    Position s=0 means it is a at ground. You find t to know the time it took to reach the ground. You then used the velocity function by taking derivative of position. Plug that time it reached the ground to the velocity function.

  13. ganeshie8
    • one year ago
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    nope, whats antiderivative of constant ?

  14. anonymous
    • one year ago
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    well you add a variable in this case t

  15. anonymous
    • one year ago
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    oh woops i see what i did wrong

  16. anonymous
    • one year ago
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    v(t) = -32t + 40: also @Shalante if we take your s(t) and we derive it until we get a(t) we get -64 and i think it should be -32

  17. anonymous
    • one year ago
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    no it is -64

  18. anonymous
    • one year ago
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    \[\int\limits v(t) = s(t) = -16t^2 + 40t + 500\] does this look right to you ?

  19. anonymous
    • one year ago
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    "se acceleration due to gravity as –32 feet per second squared"

  20. anonymous
    • one year ago
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    use *

  21. anonymous
    • one year ago
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    Actually, there lots of ways to do this. There is like 3

  22. anonymous
    • one year ago
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    yes that is right.

  23. anonymous
    • one year ago
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    i know but i have to show my work for this one so cant use physics equations sadly :(

  24. anonymous
    • one year ago
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    oh wait it is wrong. Supposed to be -32t^2-40t+500

  25. anonymous
    • one year ago
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    Make the position=0 by doing quadratic formula.

  26. anonymous
    • one year ago
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    use your calculator divide everything by -4 first

  27. Empty
    • one year ago
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    Yeah as far as physics is concerned I think this would be pretty much the best way: \[K_i + V_i = K_f + V_f\] \[\frac{1}{2} m 40^2 + m 32*500 = \frac{1}{2} m v^2\] \[\sqrt{40^2 + 2*32*500} = v\]

  28. ganeshie8
    • one year ago
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    careful, indefinite integral is unique only upto a constant

  29. anonymous
    • one year ago
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    interesting, i'm getting positive 40t because of the v(t) - v(0) part (when we added the v(0) to isolate the v(t))

  30. ganeshie8
    • one year ago
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    \[s(t)-s(0)=\int\limits_0^t v(t) -16t^2 + 40t \] so you're right, you do get \(s(t) = -16t^2+40t+500\)

  31. anonymous
    • one year ago
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    okay so now just set s(t) = 0 to find when it touches the ground right?

  32. anonymous
    • one year ago
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    wait i thought s (t) = -16t^2 .... not -32t^2

  33. ganeshie8
    • one year ago
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    physicists are overjoyed by the conservation of "anything", but i think math professors still want to see the integrals mess ;)

  34. anonymous
    • one year ago
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    Answer is negative because the ball's speed while hitting the ground is going downwards.

  35. ganeshie8
    • one year ago
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    t=0 corresponds to s=500 right ?

  36. ganeshie8
    • one year ago
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    recall that \(s(t)\) is the displacement from \(ground\) after \(t\) seconds

  37. anonymous
    • one year ago
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    \[-32t^2+40t+500=0 \] got \[t=3.376s,-4.626s\] Time should only be positive so use t=3.376s \[v(t)=-64t-40\] \[v(3.376)=-256.06 m/s\] are you sure this is right @ganeshie8

  38. anonymous
    • one year ago
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    shouldn't it be \[-16t^2 +40t + 500 = 0 \]

  39. ganeshie8
    • one year ago
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    sry im not following the discussion between you and Jdosio... but you're not allowed to use direct formulas here right ?

  40. anonymous
    • one year ago
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    no only calculus

  41. anonymous
    • one year ago
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    okay my only positive root for v(t) is 6.978

  42. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=-16x%5E2+%2B+40x+%2B+500

  43. anonymous
    • one year ago
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    Ya use -16t^2

  44. anonymous
    • one year ago
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    so to find the velocity at that time we just plug that into the formula for v(t) right ?

  45. anonymous
    • one year ago
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    which gives me around -183.3030

  46. anonymous
    • one year ago
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    does that look about right to you guys?

  47. anonymous
    • one year ago
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    Nope it should be -40

  48. anonymous
    • one year ago
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    im using v(t) = -32t + 40 is that right?

  49. anonymous
    • one year ago
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    Answer is right, but the graph is wrong. Guess you do not need the graph.

  50. anonymous
    • one year ago
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    okay so then does everyone agree that the answer is -183.3030 !?

  51. anonymous
    • one year ago
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    Read it wrong at the beginning lol

  52. anonymous
    • one year ago
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    @Empty @ganeshie8 ?

  53. anonymous
    • one year ago
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    It is correct.

  54. anonymous
    • one year ago
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    okay ill stick around just incase anyone changes their mind, thanks for the help guys :D

  55. Empty
    • one year ago
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    I got -183.3030 by doing it with the energy considerations alone, so good job! Same answer different route is a good indicator imo.

  56. ganeshie8
    • one year ago
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    Yes the ball touches the ground after at `t=6.978` \(v(6.978) =40-32(6.978) \approx -183.303\) you want to avoid that negative sign though speed is \(183.303\) whatever unit

  57. Empty
    • one year ago
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    Only reason why I think they'd want the negative sign there is because they gave -32 f/s^2 as the acceleration due to gravity, which imo is kind of awkward so I don't know what they want.

  58. ganeshie8
    • one year ago
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    acceleration is a vector, so sign makes sense speed isn't

  59. anonymous
    • one year ago
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    ill just make it explicit that the velocity is downwards

  60. ganeshie8
    • one year ago
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    they are not asking velocity

  61. Empty
    • one year ago
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    Yeah you're right @ganeshie8 good point.

  62. ganeshie8
    • one year ago
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    just persisting because, when you use the terms correctly, your professor knows that you understood the concepts.. that is for @Jdosio :)

  63. anonymous
    • one year ago
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    ill take note of that :)

  64. anonymous
    • one year ago
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    okay something weird has happened

  65. anonymous
    • one year ago
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    i found the speed of the ball at the time that the ball is at a height of 500 (the height of the cliff) and got -40 does that look oddly round to you guys ?

  66. anonymous
    • one year ago
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    @ganeshie8 @Empty @Shalante ?

  67. Empty
    • one year ago
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    Well this is actually completely expected that the speed is the same for the same height, this is because it has no more potential or kinetic energy at this same height when you threw it. This is due to conservation of energy, which I realize is not exactly in the scope of what you're learning however it's clear at a glance to me because of this.

  68. anonymous
    • one year ago
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    okay so i should stick with my 183.3030 right?

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