The range of a projectile launched at an angle of `15^(@)` with horizontal is 1.5km. The range of projectile when launched at an angle of `45^(@)` to the horizontal is

Options

A. 1.5 km
B. 3 km
C. 6km
D. 0.75km

Explanation

To solve the problem, we will use the formula for the range of a projectile, which is given by:

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

where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the given data for the first case:**
- Angle of projection \( \theta_1 = 15^\circ \)
- Range \( R_1 = 1.5 \text{ km} = 1500 \text{ m} \)

2. **Write the range formula for the first case:**
\[ R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} \]
Simplifying the sine term:
\[ \sin(30^\circ) = \frac{1}{2} \]
Therefore, we can write:
\[ R_1 = \frac{u^2 \cdot \frac{1}{2}}{g} = \frac{u^2}{2g} \]

3. **Set up the equation using the known range:**
\[ 1500 = \frac{u^2}{2g} \]
Rearranging gives:
\[ u^2 = 3000g \]

4. **Now consider the second case with \( \theta_2 = 45^\circ \):**
\[ R_2 = \frac{u^2 \sin(2 \times 45^\circ)}{g} \]
Simplifying the sine term:
\[ \sin(90^\circ) = 1 \]
Therefore, we can write:
\[ R_2 = \frac{u^2}{g} \]

5. **Substituting \( u^2 \) from the first case into the second case:**
\[ R_2 = \frac{3000g}{g} = 3000 \text{ m} = 3 \text{ km} \]

6. **Final Answer:**
The range of the projectile when launched at an angle of \( 45^\circ \) is \( 3 \text{ km} \).