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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
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@saifoo.khan

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Other answers:

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