A) climb with an acceleration 9.8\[m/{{s}^{2}}\]
B) fall with an acceleration 9.8 \[m/{{s}^{2}}\]
C) fall with a constant acceleration 3.4\[m/{{s}^{2}}\]
D) fall with acceleration and then would attain a constant velocity
Correct Answer: C
Solution :
If B is up thrust of air on balloon, and a is downward acceleration, then \[Mg-B=Ma\] \[\begin{align} & \Rightarrow a=\frac{Mg-B}{M}=g-\frac{{{V}_{pair}}g}{{{V}_{Pco}}_{_{2}}} \\ & =\left( 1-\frac{{{V}_{pair}}}{{{V}_{Pc{{o}_{2}}}}} \right)g=\left( 1-\frac{28.8}{44} \right)\times 9.8m/{{s}^{2}} \\ & =3.4m/{{s}^{2}} \\ \end{align}\]
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