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calculusxy

  • one year ago

MEDAL!!! A ball is thrown straight up. What will be the instantaneous velocity at the top of its path? What will be its acceleration at the top? Why are your answers different?

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  1. calculusxy
    • one year ago
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    @jim_thompson5910

  2. anonymous
    • one year ago
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    Is it not that when it goes upward, at the top of its path, velocity =0?

  3. jim_thompson5910
    • one year ago
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    OOOPS is correct Gravity is pulling the ball down. The initial throw pushes the ball up and the velocity decreases as the ball climbs higher because it's fighting gravity. Once the ball gets to the peak, the velocity hits 0. The acceleration due to gravity is always the same no matter how fast the ball is moving.

  4. calculusxy
    • one year ago
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    What would be its acceleration?

  5. jim_thompson5910
    • one year ago
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    The acceleration due to gravity is always the same no matter how fast the ball is moving. acceleration of gravity = 9.8 m/s^2 = 32 ft/s^2 (both are approximate)

  6. calculusxy
    • one year ago
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    How did you calculate that 9.8m/s^2 = 38ft?s^2 ?

  7. jim_thompson5910
    • one year ago
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    the acceleration of gravity only changes when the height is drastically different technically there is change, but it's so small that it's hard to notice

  8. jim_thompson5910
    • one year ago
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    it's just something you look up in a table, book, or something. or you memorize it

  9. calculusxy
    • one year ago
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    I mean i know that 9.8m/s^2 is the instantaneous speed of a free-falling object. but how did you just calculate the acceleration at the top? acceleration = gravity x time ?

  10. jim_thompson5910
    • one year ago
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    the acceleration is equal to the acceleration of gravity

  11. calculusxy
    • one year ago
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    Honestly, i have no idea to what you just said.

  12. jim_thompson5910
    • one year ago
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    imagine the object in free fall every second, the speed increases by 9.8 m/s

  13. calculusxy
    • one year ago
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    okay

  14. jim_thompson5910
    • one year ago
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    let t be the time elapsed since dropping the object and letting it free fall (ignore air resistance) t = 0 speed = 0 m/s t = 1 speed = 9.8 m/s t = 2 speed = 19.6 m/s (9.8 times 2 = 19.6) t = 3 speed = 29.4 m/s (9.8 times 3 = 29.4) etc etc

  15. calculusxy
    • one year ago
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    yeah...

  16. calculusxy
    • one year ago
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    so what's going to happen next?

  17. jim_thompson5910
    • one year ago
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    No matter where the object is, it's feeling the same force on it. So it undergoes the same acceleration throughout its journey going up and then coming back down

  18. zzr0ck3r
    • one year ago
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    acceleration is a vector. i.e. it has direction.

  19. calculusxy
    • one year ago
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    but my question was how did you get this: "acceleration of gravity = 9.8 m/s^2 = 32 ft/s^2 (both are approximate)"

  20. zzr0ck3r
    • one year ago
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    Isac Newton

  21. zzr0ck3r
    • one year ago
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    I think...

  22. jim_thompson5910
    • one year ago
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    Galileo discovered it long before Newton, but yeah pretty much

  23. zzr0ck3r
    • one year ago
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    did he discover the speed or just the thing?

  24. jim_thompson5910
    • one year ago
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    those two numbers are constants that you memorize or look up

  25. calculusxy
    • one year ago
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    so the acceleration is 32?

  26. jim_thompson5910
    • one year ago
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    32 ft per s^2, yes

  27. jim_thompson5910
    • one year ago
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    just the speed/acceleration I think @zzr0ck3r I don't think he actually came up with unifying theories like Newton did

  28. zzr0ck3r
    • one year ago
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    And almost got killed as a result. Ahh humans...

  29. calculusxy
    • one year ago
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    so i found this website where it said that the acceleration can just be defined as "g", i believe:

  30. zzr0ck3r
    • one year ago
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    yes. gravity is constant and we normally just say g because we are estimating it when we give an actually number.

  31. zzr0ck3r
    • one year ago
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    like \(\pi\) and 3.14. They are not the same but we use 3.14 when we need a close number.

  32. jim_thompson5910
    • one year ago
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    g = 9.8 m/s^2 or g = 32 ft/s^2 both are the same. Saying 'g' is a shorthand way to say "acceleration of gravity on earth at sea level"

  33. jim_thompson5910
    • one year ago
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    When you hear things like "the rocket is going 3 g's (idk I made up the number)" it means "the rocket is accelerating 3 times the acceleration of gravity" so the rocket's acceleration would be 29 m/s^2

  34. calculusxy
    • one year ago
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    Thank you. I have another question.

  35. phi
    • one year ago
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    If you have time and are interested, see https://www.khanacademy.org/science/physics/newton-gravitation/gravity-newtonian/v/introduction-to-gravity

  36. calculusxy
    • one year ago
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    If a salmon swims straight upward in the water fast enough to break through the surface at a speed of 5m/s, how high can it jump above the water?

  37. jim_thompson5910
    • one year ago
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    Acceleration of gravity: g = 9.8 m/s^2 Initial Velocity Vo = 5 m/s Plug those values into h = -(g/2)t^2 + Vo*t We don't know anything about the angle in which it jumps, so ignore that. Assume it jumps straight up. Find the vertex to get the answer.

  38. calculusxy
    • one year ago
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    h = (-9.8/2)5^2 + 5*2

  39. jim_thompson5910
    • one year ago
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    t is unknown for now

  40. calculusxy
    • one year ago
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    Yeah that's what I was guessing

  41. jim_thompson5910
    • one year ago
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    you should get h = -4.9t^2 + 5t find the vertex of this to figure out the highest point

  42. calculusxy
    • one year ago
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    How would I figured out the vertex?

  43. calculusxy
    • one year ago
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    *figure

  44. jim_thompson5910
    • one year ago
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    h = -4.9t^2 + 5t is in the form h = at^2 + bt + c a = -4.9 b = 5 c = 0 plug the values of 'a' and 'b' into -b/(2a) to figure out the time value at the peak once you get this value, plug it into h = -4.9t^2 + 5t to find h

  45. calculusxy
    • one year ago
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    so t would be -b/(2a) ?

  46. jim_thompson5910
    • one year ago
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    the t value at the peak, yes |dw:1436315387333:dw|

  47. calculusxy
    • one year ago
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    Okay

  48. calculusxy
    • one year ago
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    -5/(2 x (-4.9))

  49. jim_thompson5910
    • one year ago
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    yes

  50. calculusxy
    • one year ago
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    -5/(-9.8) = 0.51

  51. jim_thompson5910
    • one year ago
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    then you plug t = 0.51 into h = -4.9t^2 + 5t

  52. calculusxy
    • one year ago
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    -4.9(0.51)^2 + 5(0.51) -4.9(0.2601) + 5(0.51) 1.27449 + 2.55 3.82449

  53. jim_thompson5910
    • one year ago
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    1 Attachment
  54. calculusxy
    • one year ago
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    Oh yea

  55. calculusxy
    • one year ago
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    1.27551

  56. jim_thompson5910
    • one year ago
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    Yep, 1.27551 the max height is roughly 1.27551 meters and this happens at around 0.51 seconds

  57. calculusxy
    • one year ago
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    Thank you! This is a worksheet, so can u make sure all of my answers? I will post them in another post.

  58. calculusxy
    • one year ago
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    It's like 5 questions?

  59. jim_thompson5910
    • one year ago
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    ok make a brand new post

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