A ball is thrown upward with an initial velocity V_{o} 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 instataneous velocity and \gamma is a constant). Time taken

by the ball to rise to its zenith is:

  • Option 1)

    \frac{1}{\sqrt{\gamma g}}\tan^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 2)

    \frac{1}{\sqrt{\gamma g}}\sin^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 3)

    \frac{1}{\sqrt{\gamma g}}\ln(1+\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 4)

    \frac{1}{\sqrt{2\gamma g}}\tan^{-1}(\sqrt{\frac{2\gamma }{g}}V_{o})

 

Answers (1)

 

Newton's 2nd Law -

F\propto \frac{dp}{dt}

F=\tfrac{kdp}{dt} 

F=\tfrac{d\left (mv \right )}{dt} 

F=\tfrac{m\left (dv \right )}{dt}

\frac{dv}{dt}=a

Therefore  F=ma

- wherein

K=1 in C.G.S & S.I

Force can be defined as rate of change of momentum.

 

 

f_{drag}=m\gamma v^{2}

a_{net}=\frac{dv}{dt}=-(g+\gamma v^{2})

\int_{V_o}^{0}\frac{dv}{(g+\gamma v^{2})}=\int_{0}^{t}-dt

=> \frac{1}{\gamma }\int_{V_o}^{0}\frac{dv}{(g/\gamma +v^{2})}=\int_{0}^{t}-dt

=>\frac{1}{\sqrt{\gamma g}}\tan^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})=t 


Option 1)

\frac{1}{\sqrt{\gamma g}}\tan^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

Option 2)

\frac{1}{\sqrt{\gamma g}}\sin^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

Option 3)

\frac{1}{\sqrt{\gamma g}}\ln(1+\sqrt{\frac{\gamma }{g}}V_{o})

Option 4)

\frac{1}{\sqrt{2\gamma g}}\tan^{-1}(\sqrt{\frac{2\gamma }{g}}V_{o})

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