An aerostat of mass m starts coming down with a constant acceleration Determine the ballest mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected.

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,  $$\Delta m=\frac{2mw}{g+w}$$.

## Mechanics!

#### A point traversed half the distance with a velocity v0. The remaining part of the distance was covered with velocity v1 for half the time, and with velocity v2 for the other half of the time

error: Content is protected !!