If retardation produced by air resistances to projectile is one-tenth of acceleration due to gravity, the time to reach maximum height approximately-

Options

A. Increases by 9 percent
B. Decreases by 9 percent
C. Increases by 11 percent
D. Decreases by 11 percent

Explanation

To solve the problem of finding the time to reach maximum height for a projectile with air resistance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Projectile**:
- The projectile experiences two forces: the force due to gravity (downward) and the force due to air resistance (also downward).
- The acceleration due to gravity is denoted as \( g \).
- The air resistance produces a retardation that is one-tenth of the acceleration due to gravity, which is \( \frac{g}{10} \).

2. **Calculate the Net Acceleration**:
- The total downward acceleration when considering both gravity and air resistance is:
\[ a = g + \frac{g}{10} = \frac{10g}{10} + \frac{g}{10} = \frac{11g}{10} \]

3. **Use the Kinematic Equation**:
- We know that the time to reach maximum height can be calculated using the formula:
\[ v = u - at \]
- At maximum height, the final velocity \( v = 0 \). Thus, we can rearrange the equation to solve for time \( t \):
\[ 0 = u - \left(\frac{11g}{10}\right)t \]
- Rearranging gives:
\[ t = \frac{u}{\frac{11g}{10}} = \frac{10u}{11g} \]

4. **Conclusion**:
- The time to reach maximum height, considering the retardation due to air resistance, is:
\[ t \approx \frac{10u}{11g} \]