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?

- anonymous

- chestercat

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- anonymous

@sakigirl @SmoothMath @radar

- anonymous

@saifoo.khan

- anonymous

@ujjwal

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## More answers

- anonymous

Initial velocity: +8 m/s
Initial height: 21 m
That's all the info you need.

- anonymous

initial velocity of the stone or the ballon?

- 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

21=-8t+1/2(9,81)t^2
?

- TheViper

t=d/t

- 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

my equation is wrong?

- 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

You know acceleration, initial velocity, and initial height, so that's not hard.

- anonymous

s= 1/2at^2 + ut
i learned this only...can u substitute the values /

- anonymous

That equation assumes that you start at position 0.

- anonymous

or height 0, whichever way you like to think of it.

- anonymous

we are supposd to use that equation, i dont know how to work tht fr this question, please help me

- anonymous

Nope, sorry. That equation is not enough. You legitimately cannot solve this problem with only that equation.

- 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

V^2=u^2+2as
s= (u+v)/2.t

- anonymous

@ash2326 @apoorvk

- anonymous

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

- 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

so how do we make the equation ?

- anonymous

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

- anonymous

ok

- 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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