Choosing the Correct Solution for Velocity: Balloon and Stone Experiment

[rudransh verma]

Homework Statement

A balloon starting from the surface of the earth has been ascending at a uniform velocity for 4.5 sec and a stone let fall from it reaches the ground in 7 sec. Find the velocity and height of balloon when the stone was let fall.

Relevant Equations

$s=ut+\frac12at^2$, $v=u+at$

What value of t and a should be chosen?

 

[jbriggs444]

You'll need to show some effort first.

 

[rudransh verma]

You'll need to show some effort first.

$4.5u=u4.5+\frac12a4.5^2$ $$u=u+\frac12a4.5$$
I am confused why a will be equal to -g. Doesn’t the acceleration of the balloon should be considered?

 

[Orodruin]

The balloon is not accelerating as per the problem statement. Once the stone is dropped, it is not influenced by the balloon any more.

$4.5u=u4.5+\frac12a4.5^2$ $$u=u+\frac12a4.5$$
I am confused why a will be equal to -g. Doesn’t the acceleration of the balloon should be considered?

This is just formulas with number. In order for us to be able to help you, you need to tell us why you have chosen those formulas and the values you have used.

 

[rudransh verma]

The balloon is not accelerating as per the problem statement.

It is not because the net force is zero. But why is there a g(acceleration due to gravity ) also? I don’t understand this? The balloon is not slowing down!

In order for us to be able to help you, you need to tell us why you have chosen those formulas and the values you have used.

I guess that’s the most appropriate ones. We are given t, a. So the formula for s.

 

[Orodruin]

It is not because the net force is zero

Yes, it is:

Homework Statement:: ... ascending at a uniform velocity for 4.5 sec

I guess that’s the most appropriate ones. We are given t, a. So the formula for s.

This was not my question. My question was why you applied them, to what part of the motion, and why you picked the values you picked. Also, what are the remaining unknowns representing?

Physics is not just plugging numbers into an equation. You need to understand the applicability and the reasoning involved.

 

[rudransh verma]

My question was why you applied them, to what part of the motion, and why you picked the values you picked. Also, what are the remaining unknowns representing?

$s=u4.5$ That is the displacement of the balloon. Now same time t is used on the other side also (I am not sure). $u4.5+\frac12a4.5^2$. I have already said I don’t know why a=-g when the balloon is not changing it’s velocity.
The book says this eqn but I don’t understand where is the acceleration part in question.

 

[Orodruin]

That is the displacement of the balloon.

During which phase? You need to be specific.

Now same time t is used on the other side also (I am not sure).

What does the other side represent? Why is it equal and what is it describing?

I have already said I don’t know why a=-g when the balloon is not changing it’s velocity.

Again, what is the equation describing?

 

[rudransh verma]

During which phase? You need to be specific.

When ascending .

What does the other side represent? Why is it equal and what is it describing?

I think when you find the eqn using $v=u+at$ you will see this is the 2nd eqn of motion we will achieve. I asked some time before and someone told me it’s the velocity *time(the other side of eqn). It’s describing displacement

Again, what is the equation describing?

the whole eqn is actually displacement =velocity *time.

 

[Orodruin]

When ascending .

So you have expressed a relationship between the displacement during ascent and the balloon velocity. Very good. Those are your sought quantities. However, with two unknowns you need two independent relationships to solve for both.

This brings us to:

$4.5u=u4.5+\frac12a4.5^2$

What does the right hand side represent and why is it equal to the left hand side?

 

[rudransh verma]

What does the right hand side represent and why is it equal to the left hand side?

I said it It’s velocity*time which is displacement.

 

[Orodruin]

I said it It’s velocity*time which is displacement.

No, you are still talking about the left hand side. When is
$$
s = ut + at^2/2
$$
applicable in this problem?

 

[haruspex]

I don’t know why a=-g when the balloon is not changing it’s velocity.

What makes you think it is -g? You agree it is not accelerating, so it is zero. What do you notice when you plug that into your equation?

 

[bob012345]

It is not because the net force is zero. But why is there a g(acceleration due to gravity ) also? I don’t understand this? The balloon is not slowing down!

Gravity is always operating but it is the net force that determines the acceleration. In this case, the lift of the balloon exactly balances the force of gravity so the balloon rises at a constant velocity. Then the stone drops and that is a completely new problem for both the stone and the balloon. But you only care about the stone and you know the initial conditions.

 

[haruspex]

But why is there a g(acceleration due to gravity ) also?

I don’t know why a=-g when the balloon is not changing it’s velocity.
The book says this eqn

It seems to me that what is puzzling you is the official solution in a book. What is leading everyone else on this thread to be puzzled is that you have not provided the relevant extract from the book, so we have no way to understand what you are misinterpreting.

 

[kuruman]

Ahem, velocity has dimensions of length / time. Since no units of length are given explicitly or implicitly through a given value for the acceleration of gravity, the given number of 4.5 sec is not sufficient to lead to a numerical answer for the height of the balloon when the stone is let fall.

However, the velocity of the balloon (we are told) is uniform, therefore when the stone is let fall, the balloon will still be ascending with a velocity of 4.5 sec. Did I give away too much?

Edit: I misread the problem so all this is nonsense. See #21.

 

[haruspex]

Ahem, velocity has dimensions of length / time. Since no units of length are given explicitly or implicitly through a given value for the acceleration of gravity, the given number of 4.5 sec is not sufficient to lead to a numerical answer.

But if we assume a standard value for g (whatever the student has normally been told to use) there is enough info, right?

 

[bob012345]

But if we assume a standard value for g (whatever the student has normally been told to use) there is enough info, right?

Yes, this is a problem with two equations and two unknowns.

 

[kuruman]

But if we assume a standard value for g (whatever the student has normally been told to use) there is enough info, right?

Right, that's what I meant by "implicitly given" for the acceleration of gravity. My main point is that the student ought to exercise good judgment when writing numbers down and ensure that the units are consistent with the dimensions. It's a good habit to acquire, especially when one mixes numbers and symbols in an algebraic equation in which case including the units with the numbers is essential in order to sort out what's what.

As with a large majority of homework problems, this one can be first solved algebraically with the numerical substitution coming at the very end. This particular OP, despite past recommendations and exhortations, has resisted using algebraic manipulation and does not leave numerical substitution for last. If that's what OP insists on, that's fine as long as OP understands that the units must be handled correctly.

 

[bob012345]

I misread this at first but it says the balloon is ascending at a uniform velocity for 4.5 sec. I first thought it said the velocity was 4.5 m/s but it is some unknown velocity for 4.5 sec.

 

[kuruman]

I misread this at first but it says the balloon is ascending at a uniform velocity for 4.5 sec. I first thought it said the velocity was 4.5 m/s but it is some unknown velocity for 4.5 sec.

You are not alone. I misread the "for" in front of 4.5 sec as an "of" . However, my other comments in post #19 about not substituting numbers until the very end stand.

 

[SammyS]

But if we assume a standard value for g (whatever the student has normally been told to use) there is enough info, right?

Right !

Yes, this is a problem with two equations and two unknowns.

I think of it as one equation with one unknown.

Two different segments of the journey, each with a different, but known acceleration and each taking a known amount of time. The overall displacement seems to be clearly indicated (as being zero).

 

[Orodruin]

I think of it as one equation with one unknown.

”Find the velocity and height of balloon when the stone was let fall.” to me is explicitly two unknowns, but that is just splitting hairs. The relevant thing is to have the OP understand what is going on.

 

[rudransh verma]

The relevant thing is to have the OP understand what is going on.

Thank you Everyone for so many comments.
I too think the net force is zero. So no acceleration. So no suvat eqn. But in my book they have used a=-g. So I am confused.
The solution is like this: $4.5u=7u-\frac12g7^2$.
Frankly I don’t know which formula to use. But I have learned that if out of 5 variables(s,t,v,u,a)3 are given then use that eqn to find 4th.

 

[haruspex]

Thank you Everyone for so many comments.
I too think the net force is zero. So no acceleration. So no suvat eqn. But in my book they have used a=-g. So I am confused.
The solution is like this: $4.5u=7u-\frac12g7^2$.

That is for the falling stone, not for the rising balloon.
We know it was released at height (4.5s)u, at an upward speed of u, and will fall back to the ground under gravity in 7s.

 

[Orodruin]

That is for the falling stone, not for the rising balloon.
We know it was released at height (4.5s)u, at an upward speed of u, and will fall back to the ground under gravity in 7s.

I would however argue that the given solution

$4.5u=7u-\frac12g7^2$.

is still wrong.

 

[rudransh verma]

I would however argue that the given solution

I am sorry. Its $-4.5u=7u-\frac12g7^2$.
$u=20.88 m/s$.
$h=94m$.

 

[Orodruin]

I am sorry. Its $-4.5u=7u-\frac12g7^2$.
$u=20.88 m/s$.
$h=94m$.

So, do you understand why this is the case?

 

[rudransh verma]

So, do you understand why this is the case?

Absolutely not. Someone said to post the official answer. So I did. Please do explain!

 

[Orodruin]

Absolutely not. Someone said to post the official answer. So I did. Please do explain!

So, the $4.5u$ you have correctly identified as the height of the balloon when it releases the stone. If you consider the fall of the stone, can you express how far from the release point it has fallen after an arbitrary time $t$?

 

[rudransh verma]

If you consider the fall of the stone, can you express how far from the release point it has fallen after an arbitrary time t?

I guess when we talk about the fallling stone from the rest under gravity we use $s=-\frac12gt^2$. So this case is same $s=-\frac12gt^2$ except there is some initial speed of stone because the balloon is moving up. So $s=ut-\frac12gt^2$

 

[Orodruin]

I guess when we talk about the fallling stone from the rest under gravity we use $s=-\frac12gt^2$. So this case is same $s=-\frac12gt^2$ except there is some initial speed of stone because the balloon is moving up. So $s=ut-\frac12gt^2$

Right. The balloon and stone are moving together at the time of release so the stone as the same velocity as the balloon. What will be the value of $s$ when the stone hits the ground and at what $t$ does this happen?

 

[rudransh verma]

Right. The balloon and stone are moving together at the time of release so the stone as the same velocity as the balloon. What will be the value of $s$ when the stone hits the ground and at what $t$ does this happen?

So it takes $7 secs$ to reach ground for the stone and I guess the height of the release point is $4.5u$ by the formula $displacement = velocity * time$ but since we are talking about the stone so $-4.5u$.

 

[Orodruin]

So it takes $7 secs$ to reach ground for the stone and I guess the height of the release point is $4.5u$ by the formula $displacement = velocity * time$ but since we are talking about the stone so $-4.5u$.

Right, the stone needs to fall back the $4.5u$ that it was lifted by the balloon, hence the negative sign. Note that, since the stone is falling under acceleration, the displacement during the fall is given by the expression you deduced in #31, not velocity*time (unless you use average velocity, which you would have to compute).

So this now gives you what relationship?

 

[rudransh verma]

Right, the stone needs to fall back the $4.5u$ that it was lifted by the balloon, hence the negative sign. Note that, since the stone is falling under acceleration, the displacement during the fall is given by the expression you deduced in #31, not velocity*time (unless you use average velocity, which you would have to compute).

So this now gives you what relationship?

Ok! So the displacement is given by $s=ut-\frac12gt^2$ or $s={v_{avg}}t$. Here it’s first eqn. So the relationship is $-4.5u=7u-\frac12gt^2$. Thanks

I want to know when is this eqn ($s=ut+\frac12at^2$) generally used and what does the right side tell?