As dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vρg
Let m, V, ρHE, denote respectively the mass, volume and density of helium balloon and ρair density of air
Volume V of balloon displaces volume V of air.
So,
V(ρair − ρHE)g = ma, or,
V(ρair − ρHE)g = \(m\frac{dv}{dt}........(1)\)
Integrating equation (1) with respect to t, we have
V(ρair − ρHE)gt = mv
\(\frac{1}{2}mv^2=\frac{1}{2}m\frac{v^2}{m^2}\)(ρair − ρHE)2g2t2
\(=\frac{1}{2m}V^2\)(ρair − ρHE)2g2t2 ......(2)
If the balloon rises to a height h from h = ut + \(\frac{1}{2}\)at2
We get,
\(h=\frac{1}{2}at^2=\frac{1}{2}\frac{V(ρair − ρHE)}{m}gt^2\)
(∵ u = 0).........(3)
From Eqs. (3) and (2)
\(\frac{1}{2}mv^2=[V(ρ_{air} − ρ_{HE})g][\frac{1}{2m}V(ρ_{air} − ρ_{HE})gt^2]\)
= V(ρair − ρHE)gh
Rearranging the terms.
⇒ \(\frac{1}{2}mv^2+Vρ_{HE}gh=Vρ_{air}gh\)
Or KEballoon + PEballoon = changing in PE of air.
So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].