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zaphod

  • 2 years ago

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?

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  1. zaphod
    • 2 years ago
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    @sakigirl @SmoothMath @radar

  2. zaphod
    • 2 years ago
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    @saifoo.khan

  3. zaphod
    • 2 years ago
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    @ujjwal

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

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

  6. SmoothMath
    • 2 years ago
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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.

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

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

  9. SmoothMath
    • 2 years ago
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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.

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

  11. SmoothMath
    • 2 years ago
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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)

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

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

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

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

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

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

  18. SmoothMath
    • 2 years ago
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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.

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

  20. zaphod
    • 2 years ago
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    @ash2326 @apoorvk

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

  22. apoorvk
    • 2 years ago
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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.

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

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

  25. zaphod
    • 2 years ago
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    ok

  26. FoolAroundMath
    • 2 years ago
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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

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