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zaphod Group Title

A hot air balloon is 21m above the ground and is rising at 8ms when a sand bag is dropped from it. For how long does it take the sandbag to reach the ground?

  • 2 years ago
  • 2 years ago

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  1. zaphod Group Title
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    @sakigirl @SmoothMath @radar

    • 2 years ago
  2. zaphod Group Title
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    @saifoo.khan

    • 2 years ago
  3. zaphod Group Title
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    @ujjwal

    • 2 years ago
  4. SmoothMath Group Title
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    Initial velocity: +8 m/s Initial height: 21 m That's all the info you need.

    • 2 years ago
  5. zaphod Group Title
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    initial velocity of the stone or the ballon?

    • 2 years ago
  6. SmoothMath Group Title
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    Of the stone. The stone is being pulled along with the balloon while it is attached, so until the moment when it is dropped, their velocities will be the same.

    • 2 years ago
  7. zaphod Group Title
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    21=-8t+1/2(9,81)t^2 ?

    • 2 years ago
  8. TheViper Group Title
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    t=d/t

    • 2 years ago
  9. SmoothMath Group Title
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    height = (1/2)(acceleration)*t^2 +(velocity)*t + (Initial height) The key is to make sure everything has the right sign. Positive acceleration or velocity would mean going upward.

    • 2 years ago
  10. zaphod Group Title
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    my equation is wrong?

    • 2 years ago
  11. SmoothMath Group Title
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    Yes. First just write a general equation for height as a function of time h(t) = (1/2)(acceleration)*t^2 +(velocity)*t + (Initial height)

    • 2 years ago
  12. SmoothMath Group Title
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    You know acceleration, initial velocity, and initial height, so that's not hard.

    • 2 years ago
  13. zaphod Group Title
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    s= 1/2at^2 + ut i learned this only...can u substitute the values /

    • 2 years ago
  14. SmoothMath Group Title
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    That equation assumes that you start at position 0.

    • 2 years ago
  15. SmoothMath Group Title
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    or height 0, whichever way you like to think of it.

    • 2 years ago
  16. zaphod Group Title
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    we are supposd to use that equation, i dont know how to work tht fr this question, please help me

    • 2 years ago
  17. SmoothMath Group Title
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    Nope, sorry. That equation is not enough. You legitimately cannot solve this problem with only that equation.

    • 2 years ago
  18. SmoothMath Group Title
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    The one I've given though is THE SAME, except that it also includes initial height. I'm not sure why that bothers you so much.

    • 2 years ago
  19. zaphod Group Title
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    V^2=u^2+2as s= (u+v)/2.t

    • 2 years ago
  20. zaphod Group Title
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    @ash2326 @apoorvk

    • 2 years ago
  21. SmoothMath Group Title
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    Oh my goodness. h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. -_-

    • 2 years ago
  22. apoorvk Group Title
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    Initial velocity of the sand bag would be the same as that of the balloon, but after that, a retarding acceleration of 'g' due to the gravity would act on it.

    • 2 years ago
  23. zaphod Group Title
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    so how do we make the equation ?

    • 2 years ago
  24. SmoothMath Group Title
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    "h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. -_-"

    • 2 years ago
  25. zaphod Group Title
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    ok

    • 2 years ago
  26. FoolAroundMath Group Title
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    in \(s = ut + \frac{1}{2}at^{2} \), s is the "displacement" = final position - initial position. So, if you want to write in terms of final and initial, then you get: final position/height = ut + 0.5at^2 + initial position which is what @SmoothMath has written

    • 2 years ago
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