# 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 Option: 2 Option: 3 Option: 4

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

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