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4-1.Newton's Laws of Motion
hard
એક બોલને જમીન પરથી $V_0$ વેગથી ફેકવામાં આવે છે. બોલની ગતિ $m\gamma {v^2}$ જેટલા અવરોધક બળથી અવરોધાય છે (જ્યાં $m$ બૉલનું દળ , $v$ તાત્ક્ષણિક વેગ અને $\gamma $ અચળાંક છે). બોલને તેના શિરોબિંદુથી ઉઠવા માટે કેટલો સમય લાગશે?
A$\frac{1}{{\sqrt {\gamma g} }}\ln \left( {1 + \sqrt {\frac{\gamma }{g}} {V_0}} \right)$
B$\frac{1}{{\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{\gamma }{g}} {V_0}} \right)$
C$\frac{1}{{\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {\frac{\gamma }{g}} {V_0}} \right)$
D$\frac{1}{{\sqrt {2\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{2\gamma }{g}} {V_0}} \right)$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
– \left( {g + \gamma {v^2}} \right) = \frac{{dv}}{{dt}}\\
– gdt = \frac{g}{\gamma }\left( {\frac{{dv}}{{\frac{g}{\gamma } + {v^2}}}} \right)\\
{\rm{integrating}}\,0\, \to \,t\,and\,{V_0} \to 0:
\end{array}$
$\begin{array}{l}
– gt = – \sqrt {\frac{g}{\gamma }} {\tan ^{ – 1}}\left( {\frac{{{V_0}}}{{\sqrt {\frac{g}{\gamma }} }}} \right)\\
t = \frac{1}{{\sqrt {\gamma g} }}{\tan ^{ – 1}}\left( {\frac{{\sqrt \gamma }}{g}{V_0}} \right)
\end{array}$
– \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}$
$\begin{array}{l}
– gt = – \sqrt {\frac{g}{\gamma }} {\tan ^{ – 1}}\left( {\frac{{{V_0}}}{{\sqrt {\frac{g}{\gamma }} }}} \right)\\
t = \frac{1}{{\sqrt {\gamma g} }}{\tan ^{ – 1}}\left( {\frac{{\sqrt \gamma }}{g}{V_0}} \right)
\end{array}$
Standard 11
Physics
