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 step by step, we need to find the time it takes for a body dropped from the top of a tower to hit the ground, given that the distance covered in the last second of its journey is 9 times the distance covered in the first second.

### Step 1: Define the distances
Let:
- \( S_n \) = distance covered in \( n \) seconds
- \( S_{n-1} \) = distance covered in \( n-1 \) seconds
- \( x \) = distance covered in the first second

From the problem, we know that:
\[ S_n - S_{n-1} = 9x \]

### Step 2: Calculate the distance covered in the first second
Using the formula for distance under uniform acceleration:
\[ S = ut + \frac{1}{2}gt^2 \]
Since the body is dropped, the initial velocity \( u = 0 \). Therefore, the distance covered in the first second (\( S_1 \)) is:
\[ S_1 = 0 + \frac{1}{2}g(1^2) = \frac{1}{2}g \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ S_1 = \frac{1}{2} \times 10 \times 1^2 = 5 \, \text{m} \]
Thus, \( x = 5 \, \text{m} \).

### Step 3: Express the distance covered in \( n \) seconds
Using the same formula, the distance covered in \( n \) seconds (\( S_n \)) is:
\[ S_n = \frac{1}{2}g(n^2) = \frac{1}{2} \times 10 \times n^2 = 5n^2 \]

The distance covered in \( n-1 \) seconds (\( S_{n-1} \)) is:
\[ S_{n-1} = \frac{1}{2}g((n-1)^2) = \frac{1}{2} \times 10 \times (n-1)^2 = 5(n-1)^2 \]

### Step 4: Set up the equation
Now we can substitute \( S_n \) and \( S_{n-1} \) into the equation:
\[ S_n - S_{n-1} = 9x \]
This gives us:
\[ 5n^2 - 5(n-1)^2 = 9 \times 5 \]

### Step 5: Simplify the equation
Factor out the common term:
\[ 5(n^2 - (n^2 - 2n + 1)) = 45 \]
\[ 5(2n - 1) = 45 \]
Dividing both sides by 5:
\[ 2n - 1 = 9 \]

### Step 6: Solve for \( n \)
Add 1 to both sides:
\[ 2n = 10 \]
Now divide by 2:
\[ n = 5 \]

### Conclusion
The time taken for the body to reach the ground is \( n = 5 \) seconds.