A ball is thrown up with a certain velocity so that it reaches a height $$h$$. Find the ratio of the two different times of the ball reaching $$\frac{h}{3}$$ in both the directions.

Let the ball be projected vertically upward with initial speed $$u$$ from the point that we shall call the origin. It reaches its highest point after its velocity has become zero. At that highest point the vertical displacement from the origin is given to be $$h$$.

For a freely falling body, the kinematic relation between velocity $$v$$, initial velocity $$u$$, acceleration $$a$$ and displacement $$s$$ is stated first:

$$v^{2}=u^{2}+2as.$$

Here at the top $$v=0$$, the acceleration is that of gravity downward so $$a=-g$$, and the displacement is upward $$s=h$$. Substituting these values we get

$$0^{2}=u^{2}+2(-g)h \;\;\Longrightarrow\;\; u^{2}=2gh.$$

This result will be useful shortly. Now we need the two instants at which the ball passes the height $$\dfrac{h}{3}$$, once on the way up and again on the way down. For vertical motion with constant acceleration the displacement after time $$t$$ is given by

$$y = ut-\dfrac12 gt^{2}.$$

At the required instants the vertical coordinate $$y$$ equals $$\dfrac{h}{3}$$, so we write

$$ut-\dfrac12 gt^{2}=\dfrac{h}{3}.$$

Move every term to one side to obtain a quadratic equation in $$t$$:

$$-\dfrac12 gt^{2}+ut-\dfrac{h}{3}=0.$$

Multiplying through by $$-1$$ to keep the leading coefficient positive,

$$\dfrac12 gt^{2}-ut+\dfrac{h}{3}=0.$$

This is of the standard form $$at^{2}+bt+c=0$$ with

$$a=\dfrac{g}{2}, \qquad b=-u, \qquad c=\dfrac{h}{3}.$$

The quadratic-formula gives the two roots:

$$t=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a}.$$

Substituting the coefficients,

$$t=\dfrac{u\pm\sqrt{u^{2}-\dfrac{2gh}{3}}}{g}.$$

In the discriminant we already know $$u^{2}=2gh$$, hence

$$u^{2}-\dfrac{2gh}{3}=2gh-\dfrac{2gh}{3}=\dfrac{4gh}{3}.$$

The square-root therefore becomes

$$\sqrt{u^{2}-\dfrac{2gh}{3}}=\sqrt{\dfrac{4gh}{3}}=\dfrac{2}{\sqrt3}\sqrt{gh}.$$

Now put $$u=\sqrt{2gh}$$ and factor $$\sqrt{gh}$$ out of the numerators:

$$t=\dfrac{\sqrt{2gh}\;\pm\;\dfrac{2}{\sqrt3}\sqrt{gh}}{g} =\dfrac{\sqrt{gh}}{g}\left(\sqrt2 \;\pm\;\dfrac{2}{\sqrt3}\right).$$

Thus the earlier time (ascending) is

$$t_{1}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2-\dfrac{2}{\sqrt3}\right),$$

and the later time (descending) is

$$t_{2}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2+\dfrac{2}{\sqrt3}\right).$$

The required ratio of the two times is therefore

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt2-\dfrac{2}{\sqrt3}} {\sqrt2+\dfrac{2}{\sqrt3}}.$$

Multiply numerator and denominator by $$\sqrt3$$ to clear the fractional terms:

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt3\sqrt2-2} {\sqrt3\sqrt2+2} =\dfrac{\sqrt6-2}{\sqrt6+2}.$$

One may notice that an equivalent, and more compact, form is obtained by multiplying the previous numerator and denominator by $$\dfrac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$$, yielding

$$\dfrac{t_{1}}{t_{2}}=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}.$$

This expression matches Option C in the list provided.

Hence, the correct answer is Option C.