A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force $m k ν^{2}$ where $ν$ is its speed. The maximum height attained by the ball is:

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$\frac{1}{2 k} \ln (1 + \frac{k u^{2}}{g})$

$\frac{1}{2 k} \tan^{- 1} \frac{k u^{2}}{g}$

$\frac{1}{k} \tan^{- 1} \frac{k u^{2}}{2 g}$

$\frac{1}{k} \ln (1 + \frac{k u^{2}}{2 g})$

answer is C.

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Detailed Solution

$\begin{array}{l} F_{n e t} = - m g - m k v^{2} [B o t h g r a v i t y a n d r e s i s t i v e f o r c e d e c e l e r a t e t h e b a l l] \\ \Rightarrow v \frac{d v}{dy} = - g - k v^{2} \\ \Rightarrow \int_{u}^{0} \frac{v d v}{g + k v^{2}} = - \int_{0}^{H} dy \\ \Rightarrow H = \frac{1}{2 k} \ln (1 + \frac{{ku}^{2}}{g}) \end{array}$