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

Options

A. gH = `(n - 2)u^(2)`
B. 2 gH = `n^(2)u^(2)`
C. gH = `(n - 2)^(2)u^(2)`
D. 2 gH = `nu^(2)(n - 2)`

Explanation

To solve the problem step by step, we will derive the relationship between the height of the tower \( H \), the initial velocity \( u \), and the time factor \( n \).

### Step 1: Time to Reach the Highest Point
When a particle is thrown upwards with an initial velocity \( u \), it will reach its highest point when its velocity becomes zero. Using the kinematic equation:

\[ v = u + at \]

At the highest point, \( v = 0 \) and \( a = -g \) (acceleration due to gravity). Thus, we have:

\[ 0 = u - gt \]

From this, we can find the time \( t \) taken to reach the highest point:

\[ gt = u \quad \Rightarrow \quad t = \frac{u}{g} \]

### Step 2: Time to Hit the Ground
According to the problem, the time taken to hit the ground is \( n \) times the time taken to reach the highest point. Therefore, the time taken to hit the ground \( t' \) is:

\[ t' = n \cdot t = n \cdot \frac{u}{g} \]

### Step 3: Distance Traveled to Hit the Ground
Now, we will use the kinematic equation for the downward motion to find the relationship between \( H \), \( u \), and \( n \):

\[ s = ut' + \frac{1}{2} a (t')^2 \]

Here, \( s \) is the distance fallen (which is equal to the height of the tower \( H \)), \( u \) is the initial velocity (which is \( 0 \) for the downward motion), and \( a = -g \). Thus, we can write:

\[ H = 0 \cdot t' + \frac{1}{2} (-g) (t')^2 \]

This simplifies to:

\[ H = -\frac{1}{2} g (t')^2 \]

Substituting \( t' = n \cdot \frac{u}{g} \):

\[ H = -\frac{1}{2} g \left(n \cdot \frac{u}{g}\right)^2 \]

### Step 4: Simplifying the Equation
Now, we simplify the equation:

\[ H = -\frac{1}{2} g \cdot \frac{n^2 u^2}{g^2} \]

This gives us:

\[ H = -\frac{n^2 u^2}{2g} \]

Since height cannot be negative, we take the absolute value:

\[ H = \frac{n^2 u^2}{2g} \]

### Step 5: Final Relation
Thus, the relation between \( H \), \( u \), and \( n \) is:

\[ H = \frac{n^2 u^2}{2g} \]