JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2019
- Hard
- MCQ
A ball is thrown upward with an initial velocity \(V_0\) from the surface of the earth. The motion of the ball is affected by a drag force equal to \(m\gamma {v^2}\) (where \(m\) is mass of the ball, \(v\) is its Instantneous velocity and \(\gamma \) is a constant). Time taken by the ball to rise to its zenith is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{1}{{\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{\gamma }{g}} {V_0}} \right)\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
- \left( {g + \gamma {v^2}} \right) = \frac{{dv}}{{dt}}\\
- gdt = \frac{g}{\gamma }\left( {\frac{{dv}}{{\frac{g}{\gamma } + {v^2}}}} \right)\\
{\rm{integrating}}\,0\, \to \,t\,and\,{V_0} \to 0:
\end{array}\)
- \left( {g + \gamma {v^2}} \right) = \frac{{dv}}{{dt}}\\
- gdt = \frac{g}{\gamma }\left( {\frac{{dv}}{{\frac{g}{\gamma } + {v^2}}}} \right)\\
{\rm{integrating}}\,0\, \to \,t\,and\,{V_0} \to 0:
\end{array}\)
\(\begin{array}{l}
- gt = - \sqrt {\frac{g}{\gamma }} {\tan ^{ - 1}}\left( {\frac{{{V_0}}}{{\sqrt {\frac{g}{\gamma }} }}} \right)\\
t = \frac{1}{{\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\frac{{\sqrt \gamma }}{g}{V_0}} \right)
\end{array}\)
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