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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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 

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ F }= mkv ^{2}- mg\) \(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=-\left[ kv ^{2}+ g \right]\) \(\Rightarrow v \cdot \frac{ dv }{ dh }=-\left[ kv ^{2}+ g \right]\) \(\Rightarrow \int_{ u }^{0} \frac{ v \cdot dv }{ kv ^{2}+ g }=-\int_{0}^{ H } dh\)…
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