A particle is falling freely under gravity. In frist `t` secnd it covers distance `d_(1)`and in the next `t` second it covers distance `d_(2)`, then `t` is given by `:`

Options

A. `sqrt((d_(2)+d_(1))/(g)`
B. `sqrt((2(d_(2)+d_(1)))/(g)`
C. `sqrt((d_(2)-d_(1))/(g)`
D. `sqrt((2(d_(2)-d_(1)))/(g)`

Explanation

To solve the problem of a particle falling freely under gravity, we need to analyze the distances covered in two different time intervals. Let's break down the solution step by step.

### Step 1: Understand the Motion
The particle is falling freely under gravity, which means it starts from rest and accelerates downwards with an acceleration equal to \( g \) (acceleration due to gravity).

### Step 2: Use the First Equation of Motion
For the first \( t \) seconds, the distance covered \( d_1 \) can be calculated using the equation of motion:
\[ d_1 = \frac{1}{2} g t^2 \]
This equation is derived from the fact that the initial velocity \( u = 0 \) and the distance covered under constant acceleration is given by \( s = ut + \frac{1}{2} a t^2 \).

### Step 3: Calculate the Distance for the Next \( t \) Seconds
For the next \( t \) seconds (from \( t \) to \( 2t \)), the total distance covered \( d_1 + d_2 \) can be calculated using the same equation:
\[ d_1 + d_2 = \frac{1}{2} g (2t)^2 \]
This simplifies to:
\[ d_1 + d_2 = \frac{1}{2} g (4t^2) = 2g t^2 \]

### Step 4: Relate \( d_1 \) and \( d_2 \)
Now we have two equations:
1. \( d_1 = \frac{1}{2} g t^2 \)
2. \( d_1 + d_2 = 2g t^2 \)

From the second equation, we can express \( d_2 \):
\[ d_2 = 2g t^2 - d_1 \]
Substituting \( d_1 \) from the first equation:
\[ d_2 = 2g t^2 - \frac{1}{2} g t^2 = \frac{4}{2} g t^2 - \frac{1}{2} g t^2 = \frac{3}{2} g t^2 \]

### Step 5: Establish the Relationship Between \( d_1 \) and \( d_2 \)
From the above, we can see:
\[ d_2 = 3 d_1 \]

### Step 6: Express \( t \) in Terms of \( d_1 \) and \( g \)
We can now express \( t \) in terms of \( d_1 \) and \( g \). From the equation for \( d_1 \):
\[ d_1 = \frac{1}{2} g t^2 \implies t^2 = \frac{2 d_1}{g} \implies t = \sqrt{\frac{2 d_1}{g}} \]

### Final Result
Thus, the time \( t \) is given by:
\[ t = \sqrt{\frac{2 d_1}{g}} \]