Question

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

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

Answer 1

Answer: (B)

$\vec{ F }= mkv ^{2}- mg$
$\vec{ a }=\frac{\vec{ F }}{ m }=-\left[ kv ^{2}+ g \right]$
$\Rightarrow v \cdot \frac{ dv }{ dh }=-\left[ kv ^{2}+ g \right]$
$\Rightarrow \int\limits_{ u }^{0} \frac{ v \cdot d v }{ kv ^{2}+ g }=-\int\limits_{0}^{ H } dh$
$\Rightarrow \frac{1}{2 K } \ln \left[ kv ^{2}+ g \right]_{ u }^{0}=- H$
$\Rightarrow \frac{1}{2 K } \ln \left[\frac{ ku ^{2}+ g }{ g }\right]= H$