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

2. zaphod

@saifoo.khan

3. zaphod

@ujjwal

4. SmoothMath

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

5. zaphod

initial velocity of the stone or the ballon?

6. SmoothMath

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

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

8. TheViper

t=d/t

9. SmoothMath

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

my equation is wrong?

11. SmoothMath

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

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

13. zaphod

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

14. SmoothMath

That equation assumes that you start at position 0.

15. SmoothMath

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

16. zaphod

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

17. SmoothMath

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

18. SmoothMath

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

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

20. zaphod

@ash2326 @apoorvk

21. SmoothMath

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

22. 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.

23. zaphod

so how do we make the equation ?

24. SmoothMath

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

25. zaphod

ok

26. 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