Choosing the Correct Solution for Velocity: Balloon and Stone Experiment

[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.