## zaphod Group Title 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? 2 years ago 2 years ago

1. zaphod Group Title

2. zaphod Group Title

@saifoo.khan

3. zaphod Group Title

@ujjwal

4. SmoothMath Group Title

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

5. zaphod Group Title

initial velocity of the stone or the ballon?

6. SmoothMath Group Title

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 Group Title

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

8. TheViper Group Title

t=d/t

9. SmoothMath Group Title

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 Group Title

my equation is wrong?

11. SmoothMath Group Title

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 Group Title

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

13. zaphod Group Title

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

14. SmoothMath Group Title

That equation assumes that you start at position 0.

15. SmoothMath Group Title

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

16. zaphod Group Title

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

17. SmoothMath Group Title

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

18. SmoothMath Group Title

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 Group Title

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

20. zaphod Group Title

@ash2326 @apoorvk

21. SmoothMath Group Title

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

22. apoorvk Group Title

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 Group Title

so how do we make the equation ?

24. SmoothMath Group Title

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

25. zaphod Group Title

ok

26. FoolAroundMath Group Title

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