A bolldropped from the top of a tower covers a distance `7x` in the last second of its journey, where `x` is the distance coverd int the first second. How much time does it take to reach to ground?.

Explanation

To solve the problem, we need to determine the time \( t \) it takes for a ball to fall from the top of a tower, given that it covers a distance of \( 7x \) in the last second of its journey, where \( x \) is the distance covered in the first second.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is dropped from a height, meaning its initial velocity \( u = 0 \).
- The distance covered in the first second is denoted as \( x \).
- The distance covered in the last second (which is the \( t \)-th second) is given as \( 7x \).

2. **Distance in the First Second**:
- The distance covered in the first second can be calculated using the formula:
\[ x = u \cdot t + \frac{1}{2} g t^2 \]
- Since \( u = 0 \) (the ball is dropped), this simplifies to:
\[ x = \frac{1}{2} g \cdot 1^2 = \frac{g}{2} \]

3. **Distance in the Last Second**:
- The distance covered in the \( t \)-th second is given by:
\[ \text{Distance in } t\text{-th second} = s_t - s_{t-1} \]
- Where \( s_t \) is the distance fallen in \( t \) seconds and \( s_{t-1} \) is the distance fallen in \( t-1 \) seconds.
- Using the formula for distance:
\[ s_t = \frac{1}{2} g t^2 \]
\[ s_{t-1} = \frac{1}{2} g (t-1)^2 \]
- Therefore, the distance in the \( t \)-th second becomes:
\[ s_t - s_{t-1} = \frac{1}{2} g t^2 - \frac{1}{2} g (t-1)^2 \]
- Simplifying this gives:
\[ s_t - s_{t-1} = \frac{1}{2} g \left( t^2 - (t^2 - 2t + 1) \right) = \frac{1}{2} g (2t - 1) = g t - \frac{g}{2} \]

4. **Setting Up the Equation**:
- We know from the problem statement that the distance covered in the last second is \( 7x \):
\[ 7x = g t - \frac{g}{2} \]
- Substituting \( x = \frac{g}{2} \) into this equation:
\[ 7 \left(\frac{g}{2}\right) = g t - \frac{g}{2} \]
- This simplifies to:
\[ \frac{7g}{2} = g t - \frac{g}{2} \]
- Rearranging gives:
\[ \frac{7g}{2} + \frac{g}{2} = g t \]
\[ \frac{8g}{2} = g t \]
\[ 4g = g t \]

5. **Solving for Time \( t \)**:
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ t = 4 \]

### Final Answer:
The time taken for the ball to reach the ground is \( t = 4 \) seconds.