Stone released from a rising balloon

[subhradeep mahata]

Homework Statement

A balloon is rising upwards with a constant velocity of 5 m/s. When it is at a height of 30 m from the ground, a stone is released from it. Calculate the time taken by the stone to reach the ground.

Relevant Equations

s=u*t+1/2(at^2)

I solved it in ground frame like this:
-30=5t-5t², and then solving for t which comes out to be 3 sec.(displacement is 30 m downwards)
Now, I am not getting the same answer when i am trying to solve it in the frame of the balloon.
W.r.t. balloon, the initial velocity is zero, displacement is still 30 m, and acceleration is -10 m/s² downwards (as the balloon itself does not have any acceleration).
So, -30 = -5t², but t is coming out to be √6 sec. But I am aware that time does not depend on frame of reference, so I should get the same answer.
Please help me out.

 

[haruspex]

W.r.t. balloon, the initial velocity is zero

Yes, but how far will the stone have to travel in that frame? What is the ground doing?

 

[subhradeep mahata]

I think that the stone should travel the entire 30 m distance that the balloon has ascended.

 

[Orodruin]

What happens with the ground in the time it takes for the stone to travel 30 m in the balloon frame?

 

[subhradeep mahata]

I am not sure, but I think the ground now should move downwards with 5 m/s relative to balloon.

 

[Orodruin]

Right. So when the stone has moved 30 m in the balloon frame, is the ground there?

 

[neilparker62]

Maybe just re-arrange your equation: $$ -30-5t=-5t^2 $$

 

[HallsofIvy]

I wouldn't calculate it like that. The motion of the balloon after the stone has been dropped is irrelevant. Rather you treat the upward motion of the balloon as the initial velocity of the stone when it is dropped.

 

[Orodruin]

I wouldn't calculate it like that. The motion of the balloon after the stone has been dropped is irrelevant. Rather you treat the upward motion of the balloon as the initial velocity of the stone when it is dropped.

This is what was done in the OP, the question was about doing the same in the balloon frame. The -5t-30 is the position of the ground in the balloon frame, which is important if you want to know when it will be colocated with the stone.

 

[neilparker62]

This is what was done in the OP, the question was about doing the same in the balloon frame. The -5t-30 is the position of the ground in the balloon frame, which is important if you want to know when it will be colocated with the stone.

And in the same frame, the stone has zero initial velocity - hence it's position is the other side of the equation $-5t^2$.