anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
@sakigirl @SmoothMath @radar
anonymous
  • anonymous
@saifoo.khan
anonymous
  • anonymous
@ujjwal

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anonymous
  • anonymous
Initial velocity: +8 m/s Initial height: 21 m That's all the info you need.
anonymous
  • anonymous
initial velocity of the stone or the ballon?
anonymous
  • anonymous
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.
anonymous
  • anonymous
21=-8t+1/2(9,81)t^2 ?
TheViper
  • TheViper
t=d/t
anonymous
  • anonymous
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.
anonymous
  • anonymous
my equation is wrong?
anonymous
  • anonymous
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)
anonymous
  • anonymous
You know acceleration, initial velocity, and initial height, so that's not hard.
anonymous
  • anonymous
s= 1/2at^2 + ut i learned this only...can u substitute the values /
anonymous
  • anonymous
That equation assumes that you start at position 0.
anonymous
  • anonymous
or height 0, whichever way you like to think of it.
anonymous
  • anonymous
we are supposd to use that equation, i dont know how to work tht fr this question, please help me
anonymous
  • anonymous
Nope, sorry. That equation is not enough. You legitimately cannot solve this problem with only that equation.
anonymous
  • anonymous
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.
anonymous
  • anonymous
V^2=u^2+2as s= (u+v)/2.t
anonymous
  • anonymous
@ash2326 @apoorvk
anonymous
  • anonymous
Oh my goodness. h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. -_-
apoorvk
  • apoorvk
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.
anonymous
  • anonymous
so how do we make the equation ?
anonymous
  • anonymous
"h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. -_-"
anonymous
  • anonymous
ok
FoolAroundMath
  • FoolAroundMath
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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