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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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Vakul

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