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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2020
  • Hard
  • MCQ

A small ball of mass \(m\) 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 

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Answer & Solution

Correct Answer

(B) \(\frac{1}{2 k } \ln \left(1+\frac{ ku ^{2}}{ g }\right)\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ F }= mkv ^{2}- mg\)

\(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=-\left[ kv ^{2}+ g \right]\)

\(\Rightarrow v \cdot \frac{ dv }{ dh }=-\left[ kv ^{2}+ g \right]\)

\(\Rightarrow \int_{ u }^{0} \frac{ v \cdot dv }{ kv ^{2}+ g }=-\int_{0}^{ H } dh\)

\(\frac{1}{2 K } \ln \left[ kv ^{2}+ g \right]_{ u }^{0}=- H\)

\(\Rightarrow \frac{1}{2 K } \ln \left[\frac{ ku ^{2}+ g }{ g }\right]= H\)

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