Question

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?

Answer 1

@sakigirl @SmoothMath @radar

Answer 2

@saifoo.khan

Answer 3

@ujjwal

Answer 4

Initial velocity: +8 m/s
Initial height: 21 m

That's all the info you need.

Answer 5

initial velocity of the stone or the ballon?

Answer 6

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.

Answer 7

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

Answer 8

t=d/t

Answer 9

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.

Answer 10

my equation is wrong?

Answer 11

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)

Answer 12

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

Answer 13

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

Answer 14

That equation assumes that you start at position 0.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

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

Answer 20

@ash2326 @apoorvk

Answer 21

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

Answer 22

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.

Answer 23

so how do we make the equation ?

Answer 24

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

Answer 25

ok

Answer 26

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