The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increases in horizontal range?

Options

A. `20%`
B. `15%`
C. `10%`
D. `5%`

Answer

To solve the problem, we need to analyze how the increase in maximum height of a projectile affects its horizontal range, given that the angle of projection remains constant.

### Step-by-Step Solution:

1. **Understand the relationship between maximum height and initial velocity**:
The maximum height \( h \) of 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. **Determine the effect of a 5% increase in height**:
If the maximum height is increased by 5%, we can express this as:
\[ h' = h + 0.05h = 1.05h \]
Substituting the expression for \( h \):
\[ h' = \frac{u'^2 \sin^2 \theta}{2g} \]
where \( u' \) is the new initial velocity after the increase in height.

3. **Set up the equation for the new height**:
Equating the two expressions for height:
\[ 1.05 \left(\frac{u^2 \sin^2 \theta}{2g}\right) = \frac{u'^2 \sin^2 \theta}{2g} \]
Since \( \sin^2 \theta \) and \( 2g \) are constant, we can simplify:
\[ 1.05 u^2 = u'^2 \]
Taking the square root of both sides gives:
\[ u' = \sqrt{1.05} u \]

4. **Calculate the percentage increase in initial velocity**:
The percentage increase in initial velocity \( \Delta u \) can be calculated as:
\[ \Delta u = u' - u = (\sqrt{1.05} - 1)u \]
The percentage increase is:
\[ \frac{\Delta u}{u} \times 100 = (\sqrt{1.05} - 1) \times 100 \]
Using the approximation \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small \( x \):
\[ \sqrt{1.05} \approx 1 + \frac{0.05}{2} = 1 + 0.025 = 1.025 \]
Thus,
\[ \Delta u \approx 0.025u \quad \Rightarrow \quad \text{Percentage Increase} \approx 2.5\% \]

5. **Relate the change in velocity to the change in range**:
The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Since \( \sin 2\theta \) and \( g \) are constants, we can say that the range is proportional to the square of the initial velocity:
\[ R \propto u^2 \]
Therefore, the percentage change in range \( \Delta R \) can be expressed as:
\[ \frac{\Delta R}{R} \times 100 = 2 \times \frac{\Delta u}{u} \times 100 \]
Substituting the percentage increase in velocity:
\[ \Delta R \approx 2 \times 2.5\% = 5\% \]

### Final Answer:
The percentage increase in horizontal range is **5%**.