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:

We begin with the forces acting while the ball is moving upward. Its weight acts downward with magnitude $$mg$$ and the given resistive force, proportional to the square of speed, also acts downward with magnitude $$mkv^{2}$$. Taking the upward direction as positive, the net force is downward, so Newton’s second law gives

$$m\,\dfrac{dv}{dt}= -\,mg \;-\; m\,k\,v^{2}.$$

We cancel the common factor $$m$$ to simplify:

$$\dfrac{dv}{dt}= -\,g \;-\; k\,v^{2}.$$

To connect velocity with height, we use the chain‐rule identity for acceleration,

$$a = \dfrac{dv}{dt}=v\,\dfrac{dv}{dy},$$

because $$\dfrac{dy}{dt}=v$$. Substituting this into the equation of motion gives

$$v\,\dfrac{dv}{dy}= -\,g \;-\; k\,v^{2}.$$

We now separate the variables $$v$$ and $$y$$ completely:

$$\dfrac{v\,dv}{g+k\,v^{2}} = -\,dy.$$

The ball starts from the ground with speed $$u$$, so at $$y=0$$ we have $$v=u$$. At the highest point the speed becomes zero, so at $$y=H$$ we have $$v=0$$. We therefore integrate between these limits.

Left-hand integral: let $$s=v^{2} \implies ds = 2v\,dv \implies v\,dv=\dfrac{ds}{2}$$. Hence

$$\int_{u}^{0}\dfrac{v\,dv}{g+k\,v^{2}} \;=\; \dfrac12\int_{u^{2}}^{0}\dfrac{ds}{g+k\,s}.$$

The standard integral $$\displaystyle\int \dfrac{ds}{g+k\,s}= \dfrac{1}{k}\ln (g+k\,s)$$ gives

$$\dfrac12\!\left[\dfrac{1}{k}\,\ln(g+k\,s)\right]_{u^{2}}^{0} \;=\; \dfrac{1}{2k}\Bigl[\ln(g)-\ln\!\bigl(g+k\,u^{2}\bigr)\Bigr].$$

This simplifies to

$$\dfrac{1}{2k}\,\ln\!\left(\dfrac{g}{g+k\,u^{2}}\right) \;=\; -\,\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right).$$

Right-hand integral:

$$\int_{0}^{H}(-\,dy)= -\,H.$$

Equating the two evaluated integrals,

$$-\,\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right)= -\,H.$$

Both sides carry the same negative sign, so we multiply by $$-1$$ and obtain the maximum height $$H$$:

$$H=\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right).$$

This matches option D.

Hence, the correct answer is Option D.