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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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zaphodBest ResponseYou've already chosen the best response.1
@sakigirl @SmoothMath @radar
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
Initial velocity: +8 m/s Initial height: 21 m That's all the info you need.
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
initial velocity of the stone or the ballon?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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)
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
You know acceleration, initial velocity, and initial height, so that's not hard.
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
s= 1/2at^2 + ut i learned this only...can u substitute the values /
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
That equation assumes that you start at position 0.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
or height 0, whichever way you like to think of it.
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
we are supposd to use that equation, i dont know how to work tht fr this question, please help me
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
Nope, sorry. That equation is not enough. You legitimately cannot solve this problem with only that equation.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
V^2=u^2+2as s= (u+v)/2.t
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
Oh my goodness. h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. _
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
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.
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
so how do we make the equation ?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
"h(t) = (1/2)(acceleration)t^2 + (initial velocity)*t +(initial height) PLUG IN 3 THINGS. _"
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.0
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
 one year ago
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