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?.

Options

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

Explanation

To solve the problem, we need to find the time it takes for a ball dropped from the top of a tower to reach the ground, given that the distance it covers in the last second of its journey is `7x`, where `x` is the distance covered in the first second.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let `t` be the total time taken to reach the ground.
- The distance covered in the first second (`x`) can be calculated using the formula for distance covered in the first second when dropped from rest:
\[ x = \frac{1}{2} g (1^2) = \frac{g}{2} \]
- The distance covered in the last second (i.e., the `t`-th second) is given as `7x`.

2. **Distance in the Last Second**:
- The distance covered in the `t`-th second is given by the formula:
\[ d_t = u + \frac{g}{2} (2t - 1) \]
- Since the ball is dropped, the initial velocity `u = 0`. Therefore, the formula simplifies to:
\[ d_t = \frac{g}{2} (2t - 1) \]
- Substituting for `d_t`, we have:
\[ d_t = \frac{g}{2} (2t - 1) = 7x = 7 \left(\frac{g}{2}\right) \]

3. **Setting Up the Equation**:
- Now we can equate the two expressions:
\[ \frac{g}{2} (2t - 1) = 7 \left(\frac{g}{2}\right) \]
- Dividing both sides by \(\frac{g}{2}\) (assuming \(g \neq 0\)):
\[ 2t - 1 = 7 \]

4. **Solving for Time (t)**:
- Rearranging the equation:
\[ 2t = 8 \implies t = 4 \]

Thus, the time taken for the ball to reach the ground is **4 seconds**.

### Final Answer:
The ball takes **4 seconds** to reach the ground.