The velocity of projection of a body is in­creased by 2%. Other factors remaining un­changed, what will be the percentage change in the maximum height attained ?

Options

A. 0.04
B. 0.02
C. 0.03
D. 0.05

Explanation

To solve the problem, we need to determine the percentage change in the maximum height attained by a projectile when the velocity of projection is increased by 2%.

### Step-by-Step Solution:

1. **Understanding the formula for 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 velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Identifying the change in velocity**:
We are given that the velocity of projection is increased by 2%. Therefore, if the original velocity is \( u \), the new velocity \( u' \) can be expressed as:
\[ u' = u + 0.02u = 1.02u \]

3. **Calculating the change in maximum height**:
The new maximum height \( H' \) with the increased velocity can be calculated as:
\[ H' = \frac{(u')^2 \sin^2 \theta}{2g} = \frac{(1.02u)^2 \sin^2 \theta}{2g} = \frac{1.0404u^2 \sin^2 \theta}{2g} \]

4. **Finding the percentage change in height**:
The original maximum height \( H \) is:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
The percentage change in height can be calculated using the formula:
\[ \text{Percentage Change} = \frac{H' - H}{H} \times 100 \]
Substituting the values of \( H' \) and \( H \):
\[ \text{Percentage Change} = \frac{\frac{1.0404u^2 \sin^2 \theta}{2g} - \frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin^2 \theta}{2g}} \times 100 \]
Simplifying this gives:
\[ = \frac{1.0404 - 1}{1} \times 100 = 0.0404 \times 100 = 4.04\% \]

5. **Conclusion**:
Therefore, the percentage change in the maximum height attained is approximately 4.04%.

### Final Answer:
The percentage change in the maximum height attained is **4%**.