Question

A small ball of mass is thrown upward with 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. $\frac{1}{2k}{\tan }^{-1}\frac{k{u}^2}{g}$
• B. $\frac{1}{2k}\ln \left(1+\frac{k{u}^2}{g}\right)$
• C. $\frac{1}{k}{\tan }^{-1}\frac{k{u}^2}{2g}$
• D. $\frac{1}{k}\ln \left(1+\frac{k{u}^2}{2g}\right)$

Answer 1

Answer: (B)

$\vec{F}=mk{v}^2-mg$
$\vec{a}=\frac{\vec{F}}{m}=-\left[k{v}^2+g\right]$
$\Rightarrow v\cdot \frac{dv}{dh}=-\left[k{v}^2+g\right]$
$\Rightarrow \underset{u}{\overset{0}{\int }}\frac{v\cdot dv}{k{v}^2+g}=-\underset{0}{\overset{H}{\int }}dh$
$\Rightarrow \frac{1}{2K}\ln {\left[k{v}^2+g\right]}_u^0=-H$
$\Rightarrow \frac{1}{2K}\ln \left[\frac{k{u}^2+g}{g}\right]=H$