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 by what percentage does the range change ?

Answer

To solve the problem step by step, we will follow the outlined approach in the transcript:

### Step 1: Write the formula for the range of a projectile
The range \( R \) of a projectile that lands at the same elevation from which it is fired is given by:
\[ R = \frac{u^2}{g} \sin(2\theta) \]
where:
- \( u \) = initial speed of projection
- \( g \) = acceleration due to gravity
- \( \theta \) = angle of projection

### Step 2: Define the changes in initial speed and angle
We are given:
- The initial speed \( u \) is increased by 1%, which can be expressed mathematically as:
\[ \frac{\Delta u}{u} \times 100 = 1 \implies \Delta u = 0.01u \]
- The angle \( \theta \) is decreased by 2%, which can be expressed as:
\[ \frac{\Delta \theta}{\theta} \times 100 = -2 \implies \Delta \theta = -0.02\theta \]

### Step 3: Take the logarithm of both sides of the range equation
Taking the natural logarithm of the range equation:
\[ \ln R = \ln\left(\frac{u^2}{g} \sin(2\theta)\right) \]
This can be simplified using properties of logarithms:
\[ \ln R = \ln(u^2) - \ln(g) + \ln(\sin(2\theta)) \]
\[ \ln R = 2\ln u - \ln g + \ln(\sin(2\theta)) \]

### Step 4: Differentiate both sides
Differentiating both sides gives:
\[ \frac{\Delta R}{R} = 2\frac{\Delta u}{u} + \frac{\Delta(\sin(2\theta))}{\sin(2\theta)} \]
Using the chain rule for \( \sin(2\theta) \):
\[ \frac{\Delta(\sin(2\theta))}{\sin(2\theta)} = 2\frac{\Delta \theta}{\theta} \cos(2\theta) \]
Thus, we have:
\[ \frac{\Delta R}{R} = 2\frac{\Delta u}{u} + 2\frac{\Delta \theta}{\theta} \cot(2\theta) \]

### Step 5: Substitute the values
Now we substitute the values we found earlier:
- \( \frac{\Delta u}{u} = 0.01 \)
- \( \frac{\Delta \theta}{\theta} = -0.02 \)
- \( \theta = 30^\circ \) implies \( 2\theta = 60^\circ \) and \( \cot(60^\circ) = \frac{1}{\sqrt{3}} \)

Substituting these values into the equation:
\[ \frac{\Delta R}{R} = 2(0.01) + 2(-0.02) \cdot \frac{1}{\sqrt{3}} \]
Calculating each term:
\[ \frac{\Delta R}{R} = 0.02 - \frac{0.04}{\sqrt{3}} \]
Calculating \( \frac{0.04}{\sqrt{3}} \):
\[ \frac{0.04}{\sqrt{3}} \approx 0.0231 \]
Thus:
\[ \frac{\Delta R}{R} = 0.02 - 0.0231 \approx -0.0031 \]

### Step 6: Convert to percentage change
To find the percentage change in range:
\[ \frac{\Delta R}{R} \times 100 \approx -0.31\% \]

### Final Answer
The percentage change in the range is approximately:
\[ \text{Percentage change in range} \approx -0.31\% \]