Choosing the Correct Solution for Velocity: Balloon and Stone Experiment

[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?

 

[Orodruin]

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 of this eqn tell?

The right hand side is displacement under constant acceleration $a$ and initial velocity $u$. Since velocity is then given by $v = u + at$, the average velocity between time 0 and $t$ is the average between $u$ and $u + at$, in other words,
$$
v_{\rm avg} = \frac{u + (u+at)}{2} = u + \frac{at}2,
$$
leading to
$$
s = v_{\rm avg} t = ut + \frac{at^2}2.
$$

 

[rudransh verma]

The right hand side is displacement under constant acceleration a and initial velocity u.

Then the first eqn tell the final velocity under constant a and initial velocity u.
And the third tell final velocity under constant acceleration and displacement and initial velocity u. Right?

 

[mIgaMiGA]

Thank you for the information, it was quite useful

 

[bob012345]

These problems always become a lot clearer if one picks an origin and direction of the positive axis and applies it consistently to all parts of the problem. Drawing a little diagram always helps me. It seems to me there is a tendency for students to ignore the setup and dive right into using the formulas before thinking through this. In this case if $s$ is the height where positive is up, the obvious choice for the ground plane is $s=0$. In general using the notation in this thread, the complete equation is $$s = s_0 + u_0t + \frac {1}{2} a {t}^2$$ Don't ignore the initial position $s_0$. It is easiest to separate this into two parts . The first part when the stone is lifted by the balloon where $0 < t < t_1$. Because $s_0=0$ and $a=0$ we have at $t_1$; $$s_1 = u_0 t_1 $$ and the second part when the stone falls where $0 < t< t_2$. Because $a=-g$ we have; $$s_2 = s_{2_0} + u _{0} t - \frac {1}{2} g {t}^2$$ Note that we reset the clock for each part. We must apply this at $t=t_2$. The key is figuring out what is $s_{2_0}$ and what is $s_2$ at $t_2$?

 

[rudransh verma]

@Orodruin what is final velocity? Using eqn $v^2=u^2+2as$ I am getting $\sqrt(435.9-1842.4)$.
Also look into my post#37

 

[Steve4Physics]

@Orodruin what is final velocity? Using eqn $v^2=u^2+2as$ I am getting $\sqrt(435.9-1842.4)$.

If I may chip in...

Presumably you are using ‘upwards is positive’.

If you want to use $v^2=u^2+2as$, note that the displacement, $s$, is negative because the final position of the stone is below its point of release.

Since $a$ is also negative ($a = -|g| = -9.8m/s^2$) that means $2as$ will be positive. So
"$\sqrt(435.9-1842.4)$”
is wrong and should be
$\sqrt(435.9+1842.4)$
(assuming your figures are correct).

Of course, the easy way is to use $v=u+at$!

 

[rudransh verma]

is wrong and should be
(435.9+1842.4)
(assuming your figures are correct).

Of course, the easy way is to use v=u+at!

But v should come -ve not +ve.

 

[Steve4Physics]

But v should come -ve not +ve.

If using $v^2=u^2+2as$ then $v=±\sqrt{u^2+2as}$. You have to choose which solution (+ or -) is physically applicable.

If using $v = u+at$ you will automtically get the correct sign for v.

 

[rudransh verma]

If using $v^2=u^2+2as$ then $v=±\sqrt{u^2+2as}$. You have to choose which solution (+ or -) is physically applicable.

If using $v = u+at$ you will automtically get the correct sign for v.

It completely slipped from my mind. Thanks.