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.
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