# Time taken by a stone to reach the ground is independent of point of dropping

A balloon is rising. A stone is dropped and the stone takes $\frac{2u}{g}$ to reach ground (Nothing is known about $u$ and $g$ is the acceleration due to gravity), irrespective if the point of dropping. Find the acceleration of the balloon as a function of time. How to do this question? I came across it somewhere on the internet and the hint was that it can be proved that $u$ is the initial velocity of the balloon. How? Also, apart from the mathematics, some insight on how such a condition can occur. Thanks

Starting out with the general $$z := h + v~t - \frac {g}{2}~t^2,$$

where

• $t \ge 0$ is the duration since the stone had been released,

• $z[~t~]$ is the remaining "height above ground" of the stone,

• $h$ is the "heigt above ground" of ballon and stone at the release, and

• $v$ is the vertical speed of the ballon at the release.

Given $z[~\frac{2~u}{g}~] := 0$ it follows a relation between $h$ and $v$:

$$0 = h + \frac{2~u~v}{g} - \frac {g}{2}~\left(\frac{2~u}{g}\right)^2,$$

$$2~\frac{u^2}{g} = h + v~\frac{2~u}{g}.$$

Now, considering the motion of the ballon until the release, we also have $$v[~\tau~] := \frac{d}{d \tau}[~h[~\tau~]~],$$

where $\tau$ is the duration since the ballon had taken off the ground;
and consequently:

$$v[~\tau~] := \frac{d}{d \tau}[~h~] = u - h~\frac{g}{2~u},$$

$$\int ~d \tau = \int \frac{dh}{u - h~\frac{g}{2~u}},$$

$$\tau = \frac{2~u}{g}~\text{Ln}[~\frac{1}{1 - h[~\tau~]~\frac{g}{2~u^2} }~] + \text{const.}.$$

Requiring $h[~\tau = 0~] = 0$ leads to

$$\tau = \frac{2~u}{g}~\text{Ln}[~\frac{1}{1 - h[~\tau~]~\frac{g}{2~u^2} }~],$$

$$h[~\tau~] = \frac{2~u^2}{g}~(1 - \text{Exp}[~-\tau~\frac{g}{2~u}~])$$

and consequently

$$v[~\tau~] = u - h~\frac{g}{2~u} = u~\text{Exp}[~-\tau~\frac{g}{2~u}~].$$

Therefore also: $v[~\tau = 0~] = u$, as required.

Find the acceleration of the balloon as a function of [its duration since take-off]

Differentiating once more:

$$a[~\tau~] := \frac{d}{d \tau}[~v~] = \frac{d}{d \tau}[~u~\text{Exp}[~-\tau~\frac{g}{2~u}~]~] = \frac{g}{2}~\text{Exp}[~-\tau~\frac{g}{2~u}~].$$

The key is this formula is not some universal truth, but information about u(t). In the formula for the position of the stone, three constants appear: h0, v0 and g. As 2u/g should result in a time, u should be a speed: v0. Eventually this should lead to a formula containing h(t) and u(t) of the ballon.