A body dropped from the top of tower covers a distance 9 x in the last second of its journey, where x is the distance covered in the first second. How much time does it take to reach the ground.

Options

A. 5 s
B. 4 s
C. 3 s
D. 6 s

Explanation

To solve the problem, we need to determine the time it takes for a body dropped from the top of a tower to reach the ground, given that it covers a distance of \(9x\) in the last second of its journey, where \(x\) is the distance covered in the first second.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The body is dropped from rest, so its initial velocity \(u = 0\).
- The distance covered in free fall can be described by the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
- Since \(u = 0\), the equation simplifies to:
\[ s = \frac{1}{2}gt^2 \]

2. **Distance Covered in the First Second**:
- For the first second (\(t = 1\)):
\[ x = \frac{1}{2}g(1^2) = \frac{g}{2} \]

3. **Distance Covered in the Last Second**:
- Let \(n\) be the total time taken to reach the ground.
- The distance covered in the last second (from \(t = n-1\) to \(t = n\)) is given by:
\[ \text{Distance in last second} = s_n - s_{n-1} \]
- Using the equation of motion:
\[ s_n = \frac{1}{2}gn^2 \quad \text{and} \quad s_{n-1} = \frac{1}{2}g(n-1)^2 \]
- Therefore, the distance covered in the last second is:
\[ s_n - s_{n-1} = \frac{1}{2}gn^2 - \frac{1}{2}g(n-1)^2 \]
- Simplifying this:
\[ = \frac{1}{2}g[n^2 - (n^2 - 2n + 1)] = \frac{1}{2}g(2n - 1) \]

4. **Setting Up the Equation**:
- According to the problem, the distance covered in the last second is \(9x\):
\[ \frac{1}{2}g(2n - 1) = 9x \]
- Substituting \(x = \frac{g}{2}\):
\[ \frac{1}{2}g(2n - 1) = 9 \left(\frac{g}{2}\right) \]
- Canceling \(\frac{g}{2}\) from both sides (assuming \(g \neq 0\)):
\[ 2n - 1 = 9 \]

5. **Solving for \(n\)**:
- Rearranging gives:
\[ 2n = 10 \quad \Rightarrow \quad n = 5 \]

### Conclusion:
The time taken for the body to reach the ground is **5 seconds**.