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Solution:

Let R be the constant upward thurst on the aerostat of mass m, coming down with a constant acceleration w. Applying Newton’s second law of motion for the aeroslat in projection form

 \( \begin{align}  & {{F}_{y}}=m{{w}_{y}} \\  & mg-R=mw\,\,\,\,\,\,\,\,\,\,(1) \\ \end{align} \)

Now, if  \( \Delta m \) be the mass, to be dumped, then using the Eq.  \( {{F}_{y}}=m{{w}_{y}} \)

 \( R-(m-\Delta m)g=(m-\Delta m)w,\,\,\,\,\,\,\,\,(2) \)

From Eqs. (1) and (2), we get,  \( \Delta m=\frac{2mw}{g+w} \).