Problem Setup:

• A particle is thrown vertically upwards from the top of a tower of height $H$ with initial speed $u$.
• The acceleration due to gravity is $g$ (acting downwards).
• Let $t_{up}$ be the time taken to reach the highest point.
• Let $t_{total}$ be the total time taken to hit the ground.
• Given: $t_{total} = n \times t_{up}$.

Calculate Time to Highest Point

At the highest point, the final velocity $v = 0$. Using the first equation of motion, $v = u + at$:

$0 = u + (-g) t_{up}$

Solving for $t_{up}$:

$t_{up} = \frac{u}{g}$

Determine Total Time to Ground

Using the given relationship $t_{total} = n \times t_{up}$:

$t_{total} = n \times \frac{u}{g} = \frac{nu}{g}$

Relate Displacement, Velocity, and Time

Consider the motion from the top of the tower until the particle hits the ground. The displacement $s$ is downwards, so $s = -H$. The initial velocity is $u$ upwards, and acceleration is $a = -g$. The time taken is $t_{total}$.

Using the second equation of motion, $s = ut + \frac{1}{2}at^2$:

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

Substitute the expressions for $t_{total}$ and $a$:

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

$ -H = \frac{nu^2}{g} - \frac{1}{2}g\left(\frac{n^2u^2}{g^2}\right) $

$ -H = \frac{nu^2}{g} - \frac{n^2u^2}{2g} $

Combine the terms on the right side:

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

Multiply both sides by $-2g$ to isolate $H$ and simplify:

$ (-H) \times (-2g) = \frac{u^2}{2g} (2n - n^2) \times (-2g) $

$ 2gH = -u^2 (2n - n^2) $

$ 2gH = u^2 (n^2 - 2n) $

Factor out $n$:

$ 2gH = nu^2 (n - 2) $

This matches option C.