A) \[\frac{1}{\sqrt{\gamma g}}{{\sin }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]
B) \[\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]
C) \[\frac{1}{\sqrt{2\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{2\gamma }{g}}{{V}_{0}} \right)\]
D) \[\frac{1}{\sqrt{\gamma g}}\ln \left( 1+\sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]
Correct Answer: B
Solution :
\[-(g+\gamma {{\text{v}}^{2}})=\frac{d\text{v}}{dt}\] \[-gdt=\frac{g}{\gamma }\left( \frac{d\text{v}}{\frac{g}{\gamma }+{{\text{v}}^{2}}} \right)\] Integrating \[0\to t\And {{V}_{0}}\to 0\]:- \[-gt=-\sqrt{\frac{g}{\gamma }}{{\tan }^{-1}}\left( \frac{{{V}_{0}}}{\sqrt{\frac{g}{\gamma }}} \right)\] \[t=\frac{1}{\sqrt{\gamma g}}{{\tan }^{-1}}\left( \sqrt{\frac{\gamma }{g}}{{V}_{0}} \right)\]
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