The friction of the air causes a vertical retardation equal to `10%` of the acceleration due to gravity. Take `g=10 m//s^(2)` the maximum height and time to reach the maximum height will be decreased by

Options

A. `9%,9%`
B. `11%,11%`
C. `9%,10%`
D. `11%,9%`

Explanation

To solve the problem, we need to determine how the maximum height and the time to reach that maximum height are affected by the air resistance, which causes a vertical retardation equal to 10% of the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Given Values:**
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
- The air resistance causes a vertical retardation of \( 10\% \) of \( g \).
- Therefore, the effective acceleration due to gravity when considering air resistance is:
\[ g_{\text{effective}} = g + \text{retardation} = g + 0.1g = 1.1g = 1.1 \times 10 = 11 \, \text{m/s}^2 \]

2. **Finding the Maximum Height Without Air Resistance:**
- The formula for the maximum height \( H \) reached by a projectile when air resistance is neglected is given by:
\[ H = \frac{u^2}{2g} \]
- Substituting the value of \( g \):
\[ H = \frac{u^2}{2 \times 10} = \frac{u^2}{20} \]

3. **Finding the Maximum Height With Air Resistance:**
- The new maximum height \( H' \) when considering air resistance is:
\[ H' = \frac{u^2}{2g_{\text{effective}}} = \frac{u^2}{2 \times 11} = \frac{u^2}{22} \]

4. **Calculating the Decrease in Height:**
- The decrease in height \( \Delta H \) can be calculated as:
\[ \Delta H = H - H' = \frac{u^2}{20} - \frac{u^2}{22} \]
- To find a common denominator:
\[ \Delta H = \frac{11u^2 - 10u^2}{220} = \frac{u^2}{220} \]
- The percentage decrease in height is given by:
\[ \frac{\Delta H}{H} \times 100 = \frac{\frac{u^2}{220}}{\frac{u^2}{20}} \times 100 = \frac{20}{220} \times 100 = \frac{100}{11} \approx 9.09\% \]

5. **Finding the Time to Reach Maximum Height Without Air Resistance:**
- The time \( t \) to reach maximum height without air resistance is given by:
\[ t = \frac{u}{g} = \frac{u}{10} \]

6. **Finding the Time to Reach Maximum Height With Air Resistance:**
- The new time \( t' \) to reach maximum height with air resistance is:
\[ t' = \frac{u}{g_{\text{effective}}} = \frac{u}{11} \]

7. **Calculating the Decrease in Time:**
- The decrease in time \( \Delta t \) is:
\[ \Delta t = t - t' = \frac{u}{10} - \frac{u}{11} \]
- Finding a common denominator:
\[ \Delta t = \frac{11u - 10u}{110} = \frac{u}{110} \]
- The percentage decrease in time is given by:
\[ \frac{\Delta t}{t} \times 100 = \frac{\frac{u}{110}}{\frac{u}{10}} \times 100 = \frac{10}{110} \times 100 = \frac{100}{11} \approx 9.09\% \]

### Final Result:
Both the maximum height and the time to reach the maximum height will be decreased by approximately \( 9\% \).