With what velocity a ball be projected vertically so that the distance covered by it in `5^(th)` second is twice the distance it covers in its `6^(th)` second `(g=10 m//s^(2))`

Options

A. `58.8 m//s`
B. `49 m//s`
C. `65 m//s`
D. `19.6 m//s`

Explanation

To solve the problem, we need to find the initial velocity (u) with which a ball must be projected vertically such that the distance covered in the 5th second is twice the distance covered in the 6th second. We will use the equations of motion for uniformly accelerated motion, where the acceleration due to gravity (g) is acting downwards.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the initial velocity (u) such that the distance covered in the 5th second (S5) is twice the distance covered in the 6th second (S6).

2. **Formula for Distance Covered in nth Second**:
The distance covered in the nth second can be calculated using the formula:
\[ S_n = u + \left(-g\right) \cdot \frac{1}{2} \cdot (2n - 1) \]
where \( g = 10 \, \text{m/s}^2 \) (the acceleration due to gravity).

3. **Calculate S5**:
For the 5th second (n = 5):
\[ S_5 = u + \left(-10\right) \cdot \frac{1}{2} \cdot (2 \cdot 5 - 1) = u - 10 \cdot \frac{1}{2} \cdot 9 = u - 45 \]

4. **Calculate S6**:
For the 6th second (n = 6):
\[ S_6 = u + \left(-10\right) \cdot \frac{1}{2} \cdot (2 \cdot 6 - 1) = u - 10 \cdot \frac{1}{2} \cdot 11 = u - 55 \]

5. **Setting Up the Equation**:
According to the problem, the distance covered in the 5th second is twice that of the 6th second:
\[ S_5 = 2 \cdot S_6 \]
Substituting the expressions for S5 and S6:
\[ u - 45 = 2(u - 55) \]

6. **Solving the Equation**:
Expanding the equation:
\[ u - 45 = 2u - 110 \]
Rearranging gives:
\[ -45 + 110 = 2u - u \]
\[ 65 = u \]

7. **Final Result**:
The initial velocity \( u \) with which the ball must be projected is:
\[ u = 65 \, \text{m/s} \]

### Conclusion:
The ball must be projected with an initial velocity of 65 m/s.