If the friction of air causes a vertical retardation equal to 10% of the acceleration due to gravity then the maximum height and time to reach the maximum height will be decreased by `(g = 10ms^(-2))`

Options

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

Explanation

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

### Step 1: Understand the Problem
We are given:
- The acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \).
- The air resistance causes a vertical retardation of \( 10\% \) of \( g \), which is \( 0.1g = 1 \, \text{m/s}^2 \).

### Step 2: Calculate Effective Acceleration
The effective acceleration acting on the projectile when considering air resistance is:
\[ g_{\text{effective}} = g + 0.1g = 1.1g = 1.1 \times 10 = 11 \, \text{m/s}^2 \]

### Step 3: Maximum Height Calculation
The formula for maximum height \( H \) without air resistance is given by:
\[ H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \]
With air resistance, the maximum height becomes:
\[ H = \frac{u^2 \sin^2 \theta}{2 \times 1.1g} = \frac{u^2 \sin^2 \theta}{2.2g} \]

### Step 4: Calculate Percentage Decrease in Height
To find the percentage decrease in height:
\[ \text{Percentage Decrease in Height} = \frac{H_{\text{max}} - H}{H_{\text{max}}} \times 100 \]
Substituting the values:
\[ = \frac{\frac{u^2 \sin^2 \theta}{2g} - \frac{u^2 \sin^2 \theta}{2.2g}}{\frac{u^2 \sin^2 \theta}{2g}} \times 100 \]
Simplifying:
\[ = \left(1 - \frac{1}{2.2}\right) \times 100 = \left(\frac{2.2 - 2}{2.2}\right) \times 100 = \frac{0.2}{2.2} \times 100 \approx 9.09\% \]

### Step 5: Time to Reach Maximum Height Calculation
The time to reach maximum height without air resistance is:
\[ t_{\text{max}} = \frac{u \sin \theta}{g} \]
With air resistance, the time becomes:
\[ t = \frac{u \sin \theta}{1.1g} \]

### Step 6: Calculate Percentage Decrease in Time
To find the percentage decrease in time:
\[ \text{Percentage Decrease in Time} = \frac{t_{\text{max}} - t}{t_{\text{max}}} \times 100 \]
Substituting the values:
\[ = \frac{\frac{u \sin \theta}{g} - \frac{u \sin \theta}{1.1g}}{\frac{u \sin \theta}{g}} \times 100 \]
Simplifying:
\[ = \left(1 - \frac{1}{1.1}\right) \times 100 = \left(\frac{1.1 - 1}{1.1}\right) \times 100 = \frac{0.1}{1.1} \times 100 \approx 9.09\% \]

### Conclusion
Both the maximum height and time to reach maximum height decrease by approximately \( 9\% \).