The range of projectile is 50 m when `theta` is inclined with
horizontal at `15^(@).` What is the range when `theta` becomes `45^(@)` :

Options

A. 400 m
B. 300 m
C. 200 m
D. 100 m

Explanation

To solve the problem, we will use the formula for the range of a projectile:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step-by-Step Solution:

1. **Identify the given values for the first case:**
- Range \( R_1 = 50 \, \text{m} \)
- Angle \( \theta_1 = 15^\circ \)

2. **Use the range formula for the first case:**
\[ R_1 = \frac{u^2 \sin 2\theta_1}{g} \]
Substituting the known values:
\[ 50 = \frac{u^2 \sin(30^\circ)}{g} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we can rewrite the equation as:
\[ 50 = \frac{u^2 \cdot \frac{1}{2}}{g} \]
This simplifies to:
\[ 50 = \frac{u^2}{2g} \]

3. **Rearranging to find \( u^2 \):**
Multiply both sides by \( 2g \):
\[ u^2 = 100g \]

4. **Now consider the second case:**
- Angle \( \theta_2 = 45^\circ \)

5. **Use the range formula for the second case:**
\[ R_2 = \frac{u^2 \sin 2\theta_2}{g} \]
Substituting the known values:
\[ R_2 = \frac{u^2 \sin(90^\circ)}{g} \]
Since \( \sin(90^\circ) = 1 \), we can rewrite the equation as:
\[ R_2 = \frac{u^2}{g} \]

6. **Substituting \( u^2 \) from the first case into the second case:**
\[ R_2 = \frac{100g}{g} \]
This simplifies to:
\[ R_2 = 100 \, \text{m} \]

### Final Answer:
The range when \( \theta \) becomes \( 45^\circ \) is \( 100 \, \text{m} \).

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