Question

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

• A. $\frac{1}{2k}{\tan }^{-1}\left(\frac{k{u}^2}{g}\right)$
• B. $\frac{1}{k}\ln (1+\frac{k{u}^2}{2g})$
• C. $\frac{1}{k}{\tan }^{-1}\frac{k{u}^2}{g}$
• D. $\frac{1}{2k}\ln (1+\frac{k{u}^2}{g})$

Answer 1

The correct option is D $\frac{1}{2k}\ln (1+\frac{k{u}^2}{g})$



From free body diagram, the net force acting on the ball,

$\Rightarrow F_{net}=-mk{v}^2-mg$

So, acceleration

$\Rightarrow a=\frac{F_{net}}{m}=-[k{v}^2+g]$

$\Rightarrow v.\frac{dv}{dh}=-[k{v}^2+g](\because a=v\frac{dv}{dh})$

Integrating both sides,

$\Rightarrow {\int }_u^0\frac{v.dv}{k{v}^2+g}={\int }_0^{h}dh$

$\Rightarrow \frac{1}{2k}\ln {[k{v}^2+g]}_u^0=h$

$\Rightarrow \frac{1}{2k}\ln \left(\frac{k{u}^2+g}{g}\right)=h$.

Hence, $\left(D\right)$ is the correct answer.