From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

Let us call the time taken by the particle to reach the highest point of its flight $$t_1$$. At that highest point the velocity becomes zero. For motion under uniform retardation we have the first-equation of motion $$v = u + at$$. Here the final velocity is $$0$$, the initial velocity is $$u$$ and the acceleration is $$-g$$ (because gravity opposes the upward motion). Hence

$$0 = u - g t_1 \; \Longrightarrow \; t_1 = \frac{u}{g}.$$

During this upward journey the particle rises a vertical distance $$s_1$$ above the top of the tower. Using the second‐equation of motion

$$s = ut + \frac{1}{2} a t^{\,2},$$

with $$u \rightarrow u,\; t \rightarrow t_1,\; a \rightarrow -g$$ we obtain

$$s_1 = u t_1 - \frac{1}{2} g t_1^{\,2}.$$

Substituting $$t_1 = u/g$$ inside, we get

$$s_1 = u\left(\frac{u}{g}\right) - \frac{1}{2} g\left(\frac{u}{g}\right)^{\!2}

= \frac{u^{2}}{g} - \frac{1}{2}\frac{u^{2}}{g}

= \frac{u^{2}}{2g}.$$

The greatest vertical height of the particle above the ground is therefore

$$H_{\text{max}} = H + s_1 = H + \frac{u^{2}}{2g}.$$

From the highest point the particle starts its downward fall with zero initial velocity. Let the time of this fall be $$t_2$$. During this descent it covers a distance $$H + \dfrac{u^{2}}{2g}$$ under constant acceleration $$g$$. Again invoking the second‐equation of motion with $$u = 0,\; a = g,\; t = t_2$$, we have

$$H + \frac{u^{2}}{2g} = \frac{1}{2} g t_2^{\,2}.$$

Rearranging,

$$t_2^{\,2} = \frac{2}{g}\!\left(H + \frac{u^{2}}{2g}\right)

= \frac{2H}{g} + \frac{u^{2}}{g^{2}},$$

$$t_2 = \sqrt{\frac{2H}{g} + \frac{u^{2}}{g^{2}}}.$$

According to the statement of the problem the total time taken to hit the ground is $$n$$ times the time to reach the highest point. The total time is $$t_1 + t_2$$. Hence we can write

$$t_1 + t_2 = n\,t_1

\;\;\Longrightarrow\;\;

t_2 = (n-1)t_1.$$

We already have $$t_1 = \dfrac{u}{g}$$, so

$$t_2 = (n-1)\frac{u}{g}.$$

But we also have an explicit expression for $$t_2$$ obtained from the descent. Equating the two values of $$t_2$$,

$$(n-1)\frac{u}{g} = \sqrt{\frac{2H}{g} + \frac{u^{2}}{g^{2}}}.$$

Squaring both sides to eliminate the square root,

$$(n-1)^{2}\frac{u^{2}}{g^{2}} = \frac{2H}{g} + \frac{u^{2}}{g^{2}}.$$

Multiplying every term by $$g^{2}$$ gives

$$(n-1)^{2}u^{2} = 2Hg + u^{2}.$$

Now let us bring the $$u^{2}$$ term on the left next to the existing $$u^{2}$$ factor:

$$(n-1)^{2}u^{2} - u^{2} = 2Hg.$$

Taking $$u^{2}$$ common,

$$\big[(n-1)^{2} - 1\big]\,u^{2} = 2Hg.$$

Expanding and simplifying the bracket,

$$(n^{2} - 2n + 1) - 1 = n^{2} - 2n,$$

so that

$$\big[n^{2} - 2n\big]\,u^{2} = 2Hg.$$

Factoring $$n$$ out of the bracket,

$$n(n-2)\,u^{2} = 2Hg.$$

Finally writing the result in a compact symmetric form,

$$2gH = n u^{2}(n - 2).$$

This is exactly the relation given in Option C.

Hence, the correct answer is Option C.