The maximum height attained by a projectile is increased by 1% by increasing its speed of projection without changing the angle of projection. Then the percentage increase in the horizontal range will be

Options

A. `2%`
B. `1%`
C. `0.5%`
D. `0.2%`

Answer

To solve the problem, we need to analyze the relationship between the maximum height attained by a projectile and its horizontal range. We will derive the necessary equations and then find the percentage increase in the horizontal range when the speed of projection is increased such that the maximum height increases by 1%.

### Step-by-Step Solution:

1. **Understanding the Maximum Height**:
The maximum height \( H \) attained by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed of projection, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Percentage Increase in Maximum Height**:
We are given that the maximum height increases by 1%. Therefore, we can express this as:
\[ \frac{dH}{H} = 0.01 \]

3. **Differentiating the Maximum Height Equation**:
To find the relationship between the change in speed \( du \) and the change in height \( dH \), we differentiate the maximum height equation:
\[ dH = \frac{d}{du} \left( \frac{u^2 \sin^2 \theta}{2g} \right) du \]
This gives:
\[ dH = \frac{2u \sin^2 \theta}{2g} du = \frac{u \sin^2 \theta}{g} du \]

4. **Finding the Change in Speed**:
Now, substituting \( dH \) into the percentage change equation:
\[ \frac{dH}{H} = \frac{u \sin^2 \theta}{g} \cdot \frac{1}{H} du \]
Substituting \( H \):
\[ \frac{dH}{H} = \frac{u \sin^2 \theta}{g} \cdot \frac{g}{u^2 \sin^2 \theta} du = \frac{du}{u} \]
Thus, we have:
\[ \frac{dH}{H} = 0.01 \implies \frac{du}{u} = \frac{1}{2} \cdot 0.01 = 0.005 \]

5. **Finding the Horizontal Range**:
The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Now, we differentiate this equation:
\[ dR = \frac{d}{du} \left( \frac{u^2 \sin 2\theta}{g} \right) du = \frac{2u \sin 2\theta}{g} du \]

6. **Calculating the Percentage Change in Range**:
The percentage change in range can be expressed as:
\[ \frac{dR}{R} = \frac{2u \sin 2\theta}{g} \cdot \frac{1}{R} du \]
Substituting \( R \):
\[ \frac{dR}{R} = \frac{2u \sin 2\theta}{g} \cdot \frac{g}{u^2 \sin 2\theta} du = \frac{2du}{u} \]
Since we found \( \frac{du}{u} = 0.005 \):
\[ \frac{dR}{R} = 2 \cdot 0.005 = 0.01 \]

7. **Final Calculation of Percentage Increase**:
Therefore, the percentage increase in the horizontal range is:
\[ \text{Percentage Increase in Range} = 0.01 \times 100\% = 1\% \]

### Conclusion:
The percentage increase in the horizontal range when the speed of projection is increased such that the maximum height increases by 1% is **1%**.