Let m = Mass of balloon
$\begin{array}{l}\mathrm{V}=\mathrm{Volume} \mathrm{of} \mathrm{balloon}\\ {\mathrm{\rho }}_{\mathrm{He}}=\mathrm{Density} \mathrm{of} \mathrm{helium}\\ {\mathrm{\rho }}_{\mathrm{air}}=\mathrm{Density} \mathrm{of} \mathrm{air}\end{array}$
Volume V of balloon displaces volume V of air.

$\therefore$The balloon experience as upthrust

$\mathrm{V}\left({\mathrm{\rho }}_{\mathrm{air}}-{\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{g} = \mathrm{ma} = \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}$...............(i)

Integrating with respect to t,
$\begin{array}{l} \mathrm{V}\left({\mathrm{\rho }}_{\mathrm{air}}-{\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{gt}=\mathrm{mv}\end{array}$.................(ii)

From eq. (i), $\mathrm{a} = \frac{\mathrm{V}\left({\mathrm{\rho }}_{\mathrm{air}}-{\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{g}}{\mathrm{m}}$

Step 2: Find the height of the balloon in time t.
If the balloon rises to a height h, from $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$
$\text{ We get }\mathrm{h}=\frac{1}{2}{\mathrm{at}}^2=\frac{1}{2}\frac{\mathrm{V}({\mathrm{\rho }}_{\text{ air }}-{\mathrm{\rho }}_{\mathrm{He}})}{\mathrm{m}}{\mathrm{gt}}^2$.................(iii)

Step 3: Find K.E. of the balloon in time t.

$\mathrm{K}.\mathrm{E}. = \frac{1}{2}{\mathrm{mv}}^2 = \frac{{\left(\mathrm{mv}\right)}^2}{2\mathrm{m}} = \frac{{\left({\mathrm{\rho }}_{\mathrm{air}} - {\mathrm{\rho }}_{\mathrm{He}}\right)}^2{\mathrm{V}}^2{\mathrm{g}}^2{\mathrm{t}}^2}{2\mathrm{m}} \mathrm{using} \mathrm{eq}. \left(\mathrm{ii}\right)$

Step 4: Rearranging the kinetic energy in terms of potential energy.
Rearranging the terms,
$\mathrm{As} \mathrm{K}.\mathrm{E}. = \frac{1}{2}{\mathrm{mv}}^2 = \frac{{\left(\mathrm{mv}\right)}^2}{2\mathrm{m}} = \frac{{\left({\mathrm{\rho }}_{\mathrm{air}} - {\mathrm{\rho }}_{\mathrm{He}}\right)}^2{\mathrm{V}}^2{\mathrm{g}}^2{\mathrm{t}}^2}{2\mathrm{m}}$
$= \mathrm{V}\left({\mathrm{\rho }}_{\mathrm{air}} - {\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{g}\left[\frac{\left({\mathrm{\rho }}_{\mathrm{air}} - {\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{V}}{2\mathrm{m}}{\mathrm{gt}}^2\right]$
$= \mathrm{V}\left({\mathrm{\rho }}_{\mathrm{air}} - {\mathrm{\rho }}_{\mathrm{He}}\right)\mathrm{gh}$
$\Rightarrow \frac{1}{2}{\mathrm{mv}}^2+{\mathrm{V\rho }}_{\mathrm{He}}\mathrm{gh} = {\mathrm{V\rho }}_{\mathrm{air}}\mathrm{gh}$
$\Rightarrow {\mathrm{KE}}_{\mathrm{balloon}}+{\mathrm{PE}}_{\mathrm{baloon}} = \mathrm{Change} \mathrm{in} \mathrm{PE} \mathrm{of} \mathrm{air}.$

So, as the balloon goes up, an equal volume of air comes down. Increase in P.E. and K.E. of the balloon is at the cost of P.E. of air [which comes down],