The range for a projectile that lands at the same elevation from which it is fired is given by `R = (u^(2)//g) sin 2theta`. Assume that the angle of projection `= 30^(@)`. If the initial speed of projection is increased by `1%`, while the angle of projection is decreased by `2%` then the range changes by

Options

A. `-0.3%`
B. `+ 4.3%`
C. `+ 0.65%`
D. `0.85%`

Explanation

To solve the problem of how the range of a projectile changes when the initial speed and angle of projection are altered, we will follow these steps:

### Step 1: Write down the formula for the range of a projectile.
The range \( R \) of a projectile launched at an angle \( \theta \) with an initial speed \( u \) is given by:
\[ R = \frac{u^2}{g} \sin(2\theta) \]
where \( g \) is the acceleration due to gravity.

### Step 2: Identify the initial conditions.
Given:
- Initial angle of projection \( \theta = 30^\circ \)
- Initial speed \( u \) (we will consider it as \( u \) without a specific value for now)

### Step 3: Calculate the initial range.
Using \( \theta = 30^\circ \):
\[ \sin(2\theta) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus, the initial range \( R \) becomes:
\[ R = \frac{u^2}{g} \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Determine the changes in speed and angle.
- The speed is increased by \( 1\% \), so the new speed \( u' = u + 0.01u = 1.01u \).
- The angle is decreased by \( 2\% \), so the new angle \( \theta' = 30^\circ - 0.02 \cdot 30^\circ = 30^\circ - 0.6^\circ = 29.4^\circ \).

### Step 5: Calculate the new range with the updated values.
First, calculate \( \sin(2\theta') \):
\[ \theta' = 29.4^\circ \implies 2\theta' = 58.8^\circ \]
Using a calculator or sine table:
\[ \sin(58.8^\circ) \approx 0.8572 \]
Now, substitute \( u' \) and \( \sin(2\theta') \) into the range formula:
\[ R' = \frac{(1.01u)^2}{g} \cdot \sin(58.8^\circ) = \frac{1.0201u^2}{g} \cdot 0.8572 \]
Thus,
\[ R' = \frac{1.0201u^2}{g} \cdot 0.8572 \]

### Step 6: Calculate the percentage change in range.
Now, we can find the percentage change in range:
\[ \text{Percentage Change} = \frac{R' - R}{R} \times 100 \]
Substituting \( R \) and \( R' \):
\[ R' = \frac{1.0201u^2 \cdot 0.8572}{g}, \quad R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
Thus,
\[ \text{Percentage Change} = \frac{\frac{1.0201 \cdot 0.8572}{\frac{\sqrt{3}}{2}} - 1}{1} \times 100 \]
Calculating the numerical values:
\[ \frac{1.0201 \cdot 0.8572}{\frac{\sqrt{3}}{2}} \approx \frac{0.8742}{0.8660} \approx 1.009 \]
So,
\[ \text{Percentage Change} = (1.009 - 1) \times 100 \approx 0.9\% \]

### Final Answer:
The range changes by approximately \( 0.9\% \).