Answer:
\[m=V\times d=\left( \frac{4}{3}\pi {{r}^{3}}\times d \right)\]\[\because \] \[\therefore \left( \vec{V}\times \vec{A} \right)+\overrightarrow{mg}=0\] \[VA\,\,\sin \theta =mg\Rightarrow V=\frac{mg}{A\,\,\sin \,\,\theta }\] \[{{V}_{\min }}\Rightarrow \sin \theta =1\Rightarrow \theta =90{}^\circ \] \[\therefore V=\frac{mg}{A}\]
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