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A small ball of mass m is thrown upward with velocity u from the ground. the ball experiences a resistive force mkv2 where v is its speed. The maximum height attained by the balls is:
Option: 1 \frac{1}{2k}\tan ^{-1}\frac{ku^{2}}{g}
Option: 2 \frac{1}{2}\ln \left ( 1+\frac{ku^{2}}{2g} \right )
Option: 3 \frac{1}{k}\tan ^{-1}\frac{ku^{2}}{2g}
Option: 4 \frac{1}{2k}\ln \left ( 1+\frac{ku^{2}}{g} \right )

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\begin{array}{l} \overrightarrow{\mathrm{F}}=\mathrm{mkv}^{2}-\mathrm{mg} \\ \\ \overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=-\left[\mathrm{kv}^{2}+\mathrm{g}\right] \\\\ \Rightarrow \quad \mathrm{v} \cdot \frac{\mathrm{dv}}{\mathrm{dh}}=-\left[\mathrm{kv}^{2}+\mathrm{g}\right] \\\\ \Rightarrow \int_{\mathrm{u}}^{0} \frac{\mathrm{v} \cdot \mathrm{d} \mathrm{v}}{\mathrm{kv}^{2}+\mathrm{g}}=-\int_{0}^{\mathrm{H}} \mathrm{dh} \\\\ \frac{1}{2 \mathrm{~K}} \ln \left[\mathrm{kv}^{2}+\mathrm{g}\right]_{\mathrm{u}}^{0}=-\mathrm{H} \\ \\ \Rightarrow \frac{1}{2 \mathrm{~K}} \ln \left[\frac{\mathrm{ku}^{2}+\mathrm{g}}{\mathrm{g}}\right]=\mathrm{H} \end{array}

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Deependra Verma

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