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.