The friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity, take (`g=10ms^(-2))`. The maximum height will be decreased by:

Options

A. `8%`
B. `9%`
C. `10%`
D. `11%`

Explanation

To solve the problem of how much the maximum height will be decreased due to air friction, we can follow these steps:

### Step 1: Understand the effect of air friction
The problem states that the air friction causes a vertical retardation equal to 10% of the acceleration due to gravity. Given that \( g = 10 \, \text{m/s}^2 \), we can calculate the retardation due to air friction.

**Calculation:**
\[ \text{Retardation} = 0.1 \times g = 0.1 \times 10 = 1 \, \text{m/s}^2 \]

### Step 2: Determine the effective acceleration due to gravity
The effective acceleration due to gravity when considering the retardation is the sum of the gravitational acceleration and the retardation due to air friction.

**Calculation:**
\[ g' = g + \text{Retardation} = 10 + 1 = 11 \, \text{m/s}^2 \]

### Step 3: Use the formula for maximum height
The formula for the maximum height \( H_{\text{max}} \) reached by a projectile is given by:
\[ H_{\text{max}} = \frac{U^2 \sin^2 \theta}{2g} \]
where \( U \) is the initial velocity and \( \theta \) is the angle of projection.

### Step 4: Calculate the new maximum height with the effective gravity
Using the effective gravity \( g' \):
\[ H'_{\text{max}} = \frac{U^2 \sin^2 \theta}{2g'} = \frac{U^2 \sin^2 \theta}{2 \times 11} \]

### Step 5: Compare the original and new maximum heights
Let \( H_{\text{max}} \) be the original maximum height:
\[ H_{\text{max}} = \frac{U^2 \sin^2 \theta}{2g} = \frac{U^2 \sin^2 \theta}{2 \times 10} \]

### Step 6: Find the decrease in maximum height
To find the decrease in maximum height, we calculate:
\[ \text{Decrease} = H_{\text{max}} - H'_{\text{max}} = \frac{U^2 \sin^2 \theta}{2 \times 10} - \frac{U^2 \sin^2 \theta}{2 \times 11} \]

### Step 7: Simplify the expression for decrease
Finding a common denominator:
\[ \text{Decrease} = \frac{U^2 \sin^2 \theta}{20} - \frac{U^2 \sin^2 \theta}{22} \]
\[ = U^2 \sin^2 \theta \left( \frac{1}{20} - \frac{1}{22} \right) \]
Calculating the difference:
\[ = U^2 \sin^2 \theta \left( \frac{22 - 20}{440} \right) = U^2 \sin^2 \theta \left( \frac{2}{440} \right) = \frac{U^2 \sin^2 \theta}{220} \]

### Step 8: Calculate the percentage decrease
The percentage decrease in maximum height is given by:
\[ \text{Percentage Decrease} = \left( \frac{\text{Decrease}}{H_{\text{max}}} \right) \times 100 \]
Substituting \( H_{\text{max}} \):
\[ = \left( \frac{\frac{U^2 \sin^2 \theta}{220}}{\frac{U^2 \sin^2 \theta}{20}} \right) \times 100 \]
\[ = \left( \frac{20}{220} \right) \times 100 = \frac{100}{11} \approx 9.09\% \]

### Conclusion
The maximum height will be decreased by approximately \( 9.09\% \).

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