Step 1: Understanding the Concept:
This is a problem of motion under gravity in one dimension.
The particle starts from a height \(H\), travels upwards to a maximum height, stops momentarily, and then falls back down to the ground.
We need to compare the time to reach the peak with the total time of flight.
The time to reach the peak (\(t_1\)) depends only on the initial upward velocity \(u\) and acceleration due to gravity \(g\).
The total time of flight (\(T\)) depends on the initial height \(H\), initial velocity \(u\), and gravity \(g\).
Step 2: Key Formula or Approach:
Time to reach highest point: \(v = u - gt \implies 0 = u - gt_1 \implies t_1 = \frac{u}{g}\).
Equation for displacement: \(s = ut + \frac{1}{2}at^2\).
Taking the point of projection as origin and upward as positive:
Displacement at ground, \(s = -H\).
Initial velocity, \(u = +u\).
Acceleration, \(a = -g\).
Step 3: Detailed Explanation:
First, find the time to reach the maximum height:
\[ t_1 = \frac{u}{g} \]
Given that the total time taken to hit the ground is \(T = n \cdot t_1\):
\[ T = \frac{nu}{g} \]
Now, apply the displacement equation for the entire trip from projection to hitting the ground:
\[ s = uT + \frac{1}{2}aT^2 \]
Substituting the values \(s = -H\), \(a = -g\), and \(T = \frac{nu}{g}\):
\[ -H = u\left(\frac{nu}{g}\right) - \frac{1}{2}g\left(\frac{nu}{g}\right)^2 \]
\[ -H = \frac{nu^2}{g} - \frac{1}{2}g \frac{n^2u^2}{g^2} \]
\[ -H = \frac{nu^2}{g} - \frac{n^2u^2}{2g} \]
Multiply the entire equation by \(-2g\) to clear the denominators and the negative sign:
\[ 2gH = -2nu^2 + n^2u^2 \]
\[ 2gH = n^2u^2 - 2nu^2 \]
Factoring out \(nu^2\) on the right side:
\[ 2gH = nu^2(n - 2) \]
This matches option (C).
Step 4: Final Answer:
By substituting the total time \(T = nu/g\) into the kinematic displacement equation, we derive the algebraic relationship \(2gH = nu^2(n - 2)\).