The maximum height attaine by a projectile is increased by ` 10%` by increasing its speed of projecton, without changing the angle of projection. What will the percentage increase in the horizontal range.

Options

A. 0.2
B. 0.15
C. 0.1
D. 0.05

Explanation

To solve the problem, we need to analyze the relationship between the maximum height attained by a projectile and its horizontal range, given that the speed of projection is increased by 10% without changing the angle of projection.

### Step-by-Step Solution:

1. **Understand the formulas**:
- The maximum height \( H_{max} \) of a projectile is given by:
\[ H_{max} = \frac{u_0^2 \sin^2 \theta}{2g} \]
- The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u_0^2 \sin 2\theta}{g} \]
Here, \( u_0 \) is the initial speed of projection, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Proportionality**:
- Both \( H_{max} \) and \( R \) depend on \( u_0^2 \). Therefore, if \( H_{max} \) increases, \( R \) will also increase proportionally to the square of the speed of projection.

3. **Given information**:
- The maximum height \( H_{max} \) is increased by 10%. This means:
\[ H_{max}' = H_{max} + 0.1 H_{max} = 1.1 H_{max} \]
- Since \( H_{max} \) is proportional to \( u_0^2 \), we can express this increase in terms of the initial speed:
\[ H_{max}' = \frac{(u_0')^2 \sin^2 \theta}{2g} \]
- Setting the two expressions for \( H_{max} \) equal gives:
\[ 1.1 \left(\frac{u_0^2 \sin^2 \theta}{2g}\right) = \frac{(u_0')^2 \sin^2 \theta}{2g} \]
- Simplifying, we find:
\[ (u_0')^2 = 1.1 u_0^2 \]

4. **Finding the new range**:
- The new range \( R' \) can be expressed as:
\[ R' = \frac{(u_0')^2 \sin 2\theta}{g} \]
- Substituting \( (u_0')^2 \):
\[ R' = \frac{1.1 u_0^2 \sin 2\theta}{g} \]
- The original range \( R \) is:
\[ R = \frac{u_0^2 \sin 2\theta}{g} \]
- Therefore, we can express the new range in terms of the original range:
\[ R' = 1.1 R \]

5. **Calculating the percentage increase in range**:
- The percentage increase in range is given by:
\[ \text{Percentage Increase} = \left(\frac{R' - R}{R}\right) \times 100\% \]
- Substituting \( R' \):
\[ \text{Percentage Increase} = \left(\frac{1.1 R - R}{R}\right) \times 100\% = (0.1) \times 100\% = 10\% \]

### Final Answer:
The percentage increase in the horizontal range is **10%**.