Falling stones from a balloon

Let $h(t)$ denote the height of the balloon $t$ seconds into its journey.

If we release the stone at time $t = t_0$ , then it will experience projectile motion with initial height $h(t_0)$ and initial velocity $h'(t_0)$ , giving the equation of motion for the height $f(t)$ the stone is above ground as $f(t) = h(t_0) + h'(t_0)\cdot (t-t_0) - \frac{g}{2}\cdot (t-t_0)^2, \qquad t\geq t_0$ We're told that the stone hits the ground $\frac{2u}{g}$ seconds after it is released, so $0 = f\left(t_0 + \frac{2u}{g}\right) = h(t_0) + h'(t_0)\cdot \left(\frac{2u}{g}\right) - \frac{g}{2} \cdot \left(\frac{2u}{g}\right)^2$ which can be rearranged as $h'(t_0) + \frac{g}{2u}\cdot h(t_0) = u.$

The above derivation works for every $t_0$ and $u$ is independent of $t_0$ , so it still works if we replace the arbitrary constant $t_0$ with the variable $t$ . Also, since the balloon is at ground level at $t=0$ , it follows that $h$ satisfies the following initial value problem: $\begin{aligned}h'(t) + \frac{g}{2u}\cdot h(t) &= u, \qquad t\geq 0 \\ h(0) &= 0\end{aligned}$

This may be solved by your favorite method [I'm partial to treating it as a first-order linear differential equation, if you couldn't guess by how I wrote it] to find $h(t) = \frac{2u^2}{g}\left(1 - e^{-gt/(2u)}\right)$

Then differentiating twice $h''(t) = -\frac{g}{2}e^{-gt/(2u)}$ and setting $g=10$ , $t=5$ , and $u=25$ yields $h''(5) = -5e^{-1} \approx -1.83939721\,\,\, m/s^2$

Since the question asks for the magnitude of the acceleration, the answer is $\boxed{1.839}$ .

We will first write the equation of motion of stone.Let upwards be positive.

$\large (v(t))t-\frac{g}{2}t^{2}=-y(t)$ It is given that $t=5s$ .

By differentiating this equation of motion wrt time, $\large a(t)=\frac{-v(t)}{5}$

Now,by integrating this equation of motion wrt time, $\large -5ln\frac{|v(t)|}{25}=-5ln\frac{|-a(t)|}{5}=t \implies |a(5)|=\frac{5}{e}$

We begin with the standard equation of motion : -

$h_t = h_0 + vt - \dfrac{gt^2}{2}$

where ;

$h_t$ is the height of the stone from the ground at any time $t$ .

$h_0$ is the height of the balloon from the ground at the time of dropping of the stone.

$v$ is the velocity of the balloon at the time of dropping of stone.

Putting $t = \dfrac{2u}{g}$ , we get :-

$\dfrac{2u^2}{g} =\dfrac{2u}{g}v + h_0$

Rearranging, we get :-

$v = u - \dfrac{gh_0}{2u}$

Now, the velocity of the balloon is but the rate of change of the height of the balloon above the ground, i.e. $\dfrac{dh_0}{dt}$ Thus,

$\dfrac{dh_0}{dt} = u - \dfrac{gh_0}{2u}$

$\cfrac{dh_0}{\left(u - \cfrac{gh_0}{2u}\right)} = dt$

Integrating both sides :-

$\displaystyle\int\cfrac{dh_0}{\left(u - \cfrac{gh_0}{2u}\right)} = \displaystyle\int{dt}$

NOTE the fact that we haven’t used any limits here, so we will have to introduce a constant of integration.

$T = -\dfrac{2u}{g}\ln\left(u - \dfrac{gh_0}{2u}\right) + C$

Now, it is given that at time $t = 0$ , the balloon is on the ground.( $h_0 = 0$ )

$0 = - \dfrac{2u}{g}\ln u + C$ So,

$C = \dfrac{2u}{g}\ln u$

Using this, we have : -

$T = \dfrac{2u}{g}\ln\cfrac{u}{\left(u -\cfrac{gh_0}{2u}\right)}$

Rearranging and making, $h_0$ as the subject of the equation we have :-

$\boxed{ \displaystyle h_0 = \dfrac{2u^2}{g}\left[1 - \displaystyle e^{\left(-\dfrac{gT}{2u}\right)} \right] }$

Note at this point that we STILL don’t know what the hell $u$ is.

Lets find out.

Differentiate the equation for $h_0$ w.r.t time, and we have :-

$\boxed{v_T = ue^{\left(\dfrac{-gT}{2u}\right)} }$

Now, lets put $T = 0$ .

$v_{T = 0} = \dfrac{u}{e^0}$ So,

$u = v_{T = 0}$

Thus, $u$ is the velocity with which the balloon rises from the ground.

Now to the final part of the question.

Differentiating the velocity of the balloon w.r.t time, we will get te acceleration of the balloon.

$\boxed{a_t = -\dfrac{g}{2}e^{\left(-\dfrac{gT}{2u}\right)} }$

Thus, now substituting the values of $T$ and $u$ given in the question, we have :-

$a_{T =5} = \dfrac{g}{2}e^{-1}$ So,

$\boxed{a_{T =5} = -1.839 m/s^2}$