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(B) When the ball is moving upward,both the gravitational force and the resistive force act downwards. The net force $F$ is given by $F = -(mg + mkv^{

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

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(B) When the ball is moving upward,both the gravitational force and the resistive force act downwards. The net force $F$ is given by $F = -(mg + mkv^{2})$.Using Newton's second law,the acceleration $a$ is $a = \frac{F}{m} = -(g + kv^{2})$.We know that $a = v \frac{dv}{dh}$,so $v \frac{dv}{dh} = -(g + kv^{2})$.Rearranging the terms to integrate,we get $\frac{v \, dv}{g + kv^{2}} = -dh$.Integrating from the initial velocity $u$ at $h=0$ to final velocity $0$ at $h=H$:$\int_{u}^{0} \frac{v \, dv}{g + kv^{2}} = -\int_{0}^{H} dh$.Let $I = g + kv^{2}$,then $dI = 2kv \, dv$,or $v \, dv = \frac{dI}{2k}$.Substituting this into the integral:$\frac{1}{2k} \int_{g+ku^{2}}^{g} \frac{dI}{I} = -H$.$\frac{1}{2k} [\ln I]_{g+ku^{2}}^{g} = -H$.$\frac{1}{2k} [\ln g - \ln(g + ku^{2})] = -H$.$\frac{1}{2k} \ln \left( \frac{g}{g + ku^{2}} \right) = -H$.Multiplying by $-1$,we get $H = \frac{1}{2k} \ln \left( \frac{g + ku^{2}}{g} \right) = \frac{1}{2k} \ln \left( 1 + \frac{ku^{2}}{g} \right)$.

$A$ 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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