A freely falling body takes `t` second to travel `(1//x)^(th)` distance then time of descent is

Options

A. `(t)/(sqrtx)`
B. `t sqrtx`
C. `(sqrtx)/(t)`
D. `(1)/(t sqrtx)`

Answer

To solve the problem, we need to determine the total time of descent \( T \) for a freely falling body that takes \( t \) seconds to travel a distance of \( \frac{h}{x} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A freely falling body is dropped from a height \( h \) with an initial velocity \( u = 0 \).
- It takes \( t \) seconds to travel a distance of \( \frac{h}{x} \).

2. **Using the Second Equation of Motion**:
- The second equation of motion states:
\[ s = ut + \frac{1}{2} g t^2 \]
- Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]
- For our case, the distance \( s \) is \( \frac{h}{x} \), and we can write:
\[ \frac{h}{x} = \frac{1}{2} g t^2 \quad \text{(Equation 1)} \]

3. **Finding the Total Time of Descent**:
- Let \( T \) be the total time of descent from height \( h \) to the ground.
- Using the second equation of motion again for the total distance \( h \):
\[ h = \frac{1}{2} g T^2 \quad \text{(Equation 2)} \]

4. **Dividing the Two Equations**:
- We can divide Equation 2 by Equation 1:
\[ \frac{h}{\frac{h}{x}} = \frac{\frac{1}{2} g T^2}{\frac{1}{2} g t^2} \]
- This simplifies to:
\[ x = \frac{T^2}{t^2} \]

5. **Solving for \( T \)**:
- Rearranging the equation gives:
\[ T^2 = x t^2 \]
- Taking the square root of both sides results in:
\[ T = t \sqrt{x} \]

6. **Final Answer**:
- The time of descent \( T \) is:
\[ T = t \sqrt{x} \]