# Stone released from balloon [closed]

Q: If a stone is released from a balloon rising with acceleration a at the instant when its velocity is v. Then immediately after release the acceleration and velocity of the stone are ?

I tried to use the concept of relative velocity (I am still struggling to get answers to problems on relative velocity). I wrote $\vec {v_s}$ = $\vec {v_{s,b}}$ +$\vec{ v_b}$ (s=stone b=balloon s,b= stone wrt balloon)

$\vec {v_{s,b}}$= v and $\vec{ v_b}$= 0 (I am not too sure of this, that is why the question) Which gives me $\vec {v_s}$= v (upwards)

Also I wrote, $\vec {a_s}$ = $\vec {a_{s,b}}$ +$\vec{ a_b}$

$\vec {a_{s,b}}$=g+a and $\vec{ a_b}$=g which gives me $\vec {a_s}$= 2g+a(downwards)

As per the book my answer is incorrect .I want to know where my concept is mistaken is as I am getting incorrect solutions to all problems on relative velocity.

Is there any alternate way to solve problems based on this? (just asking in case)

## closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind♦, user36790, DanielSankOct 11 '15 at 17:11

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• Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Oct 11 '15 at 11:04
• @JohnRennie This is not a homework question.Even if was, I don't think I violated any rules though. – Karan Singh Oct 11 '15 at 11:43
• I think @JohnRennie commented about homework because our policy regards questions which are homework-like. Also, the question is hard to answer because some of the symbols are not defined; what is v? – DanielSank Oct 11 '15 at 17:10
• v is the velocity and is part of the question.It doesn't need to be separately defined as it is part of the question. – Karan Singh Oct 12 '15 at 12:21

I read your question. The answer is that the stone will have upward velocity v and downward acceleration g.

$\vec v_{s,b}= v$ and $\vec v_b= 0$ (I am not too sure of this, that is why the question) Which gives me $v_s= v$ (upwards)

This is wrong.$v_s= v$ (upwards) is no doubt correct. But $\vec v_{s,b}= 0$ and $\vec v_b= v$ since when the stone is released from the balloon, no external force acts on the stone-balloon system other than gravity. So there is no relative velocity of stone w.r.t. balloon. And the balloon velocity w.r.t. ground is given to be $v$.

$\vec a_{s,b} = g+a$ and $\vec a_b = g$ which gives me $\vec a_s = 2g+a$(downwards)

This is also wrong. When the stone loses contact with the balloon, it experiences $g$ but no longer experiences the force which accelerates the balloon i.e. it no longer gets the "a" acceleration. Therefore, $\vec a_{s,b} = (-a)$. But according to given information, $\vec a_b = (a-g)$ . Hence we get the result, $\vec a_s = -g$ i.e. $g$ in the downward direction.

UPDATE after conversation with Karan Singh:

Expansion on that part of my original answer where I said relative velocity of stone with respect to balloon is 0.

See a force is surely acting on the balloon which gives it an acceleration of "a". When the stone is in the balloon system, it also experiences the same force and acceleration. But when it is thrown out, it is no longer subjected to that force or acceleration. So the difference in acceleration as I have stated above. Now for the relative velocity part..... When you let a body fall out of a moving object (accelerating or non-accelerating), the falling body retains the velocity at the instant it leaves the moving system. One thing I must make clear that wherever I have written "thrown out", it will actually be " allowed to fall" cuz while throwing one may supply an additional accln to the body but this wont happen if it is allowed to fall from rest. Moreover when the stone is in the balloon, they act as a system and both have same velocity and subject to same force. As the stone is allowed to fall out of the balloon, the stone leaves the system with the velocity it had at that instant and experiences a sudden change in accln. due to which the velocity of the stone w.r.t. balloon at that instant remains 0 and becomes non-zero the next instant only.

• Thank you for the excellent answer.Can you suggest an alternate way to approach the problem? – Karan Singh Oct 11 '15 at 11:50
• Actually this is the best method. And perhaps the only method (I think so). Try to find out which force is acting where and apply Newton's Laws, conserve momentum,energy and use kinematic equations if needed. That is the basic rule for solving these relative velocity problems. – SchrodingersCat Oct 11 '15 at 11:54
• Sorry to ask this weird question but where did you learn this? Can you help me by providing a link or a pdf somewhere online? I do want to read more on it. – Karan Singh Oct 11 '15 at 11:58
• Now thats really weird. You need to study the chapters on mechanics from any good book (Feynman, University Physics, Berkeley). As for a pdf, you can read this...... feynmanlectures.caltech.edu/I_10.html – SchrodingersCat Oct 11 '15 at 12:22
• For the second part of the answer, you said acc. of stone wrt balloon is -a.Now is it equivalent to the following situation: I am on hot air balloon currently on the ground, and as soon as I put a stone on the ground, the balloon starts rising from the ground.So relative accn of stone wrrt balloon is -a and relative velocity is (can't figure this part out)? – Karan Singh Oct 11 '15 at 12:28

With respect to the balloon the stone will have zero velocity and an acceleration of g + a downwards.Especially if the balloon is moving up.

• Can you tell me how you got that?The book's answer is different though. – Karan Singh Oct 11 '15 at 7:36
• Since the stone is released from the balloon the stone will have no acceleration except gravity, which is downwards ,and since the stone was having a velocity v upwards while on the balloon the stone will continue to travel upwards for time v/g. – Sathyaram Oct 11 '15 at 7:40
• @SathyaramGanapathy your answer is wrong. Stone will have upward velocity $v$ and downward acceleration $g$. Kindly re-consider your logic. – SchrodingersCat Oct 11 '15 at 9:42
• I think you have missed to read the words With respect to the balloon – Sathyaram Oct 11 '15 at 14:00
• Why is the logic incorrect? The same was explained by you in your answer @Aniket. And it wasn't given that the acceleration of balloon is (a-g) and w.r.t the ground the balloon was having an acceleration a upwards already ,why to subtract it with g. – Sathyaram Oct 11 '15 at 14:25