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

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)

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

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-