The range of a projectile when fired at `75^(@)` with the horizontal is 0.5km. What will be its range when fired at `45^(@)` with same speed:-

Options

A. 0.5 km.
B. 1.0 km.
C. 1.5 km.
D. 2.0 km.

Explanation

To solve the problem, we need to find the range of a projectile when fired at an angle of 45 degrees, given that its range at 75 degrees is 0.5 km. We will use the formula for the range of a projectile:

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where \( v \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

2. **Calculate the Range at 75 Degrees**:
For the first case, where the angle \( \theta = 75^\circ \):
\[ R_{75} = \frac{v^2 \sin(150^\circ)}{g} \]
We know that \( \sin(150^\circ) = \sin(30^\circ) = \frac{1}{2} \). Therefore, we can write:
\[ R_{75} = \frac{v^2 \cdot \frac{1}{2}}{g} \]
Given that \( R_{75} = 0.5 \, \text{km} \), we can set up the equation:
\[ 0.5 = \frac{v^2 \cdot \frac{1}{2}}{g} \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ v^2 = g \cdot 0.5 \]

4. **Calculate the Range at 45 Degrees**:
Now, we will calculate the range when the projectile is fired at \( 45^\circ \):
\[ R_{45} = \frac{v^2 \sin(90^\circ)}{g} \]
Since \( \sin(90^\circ) = 1 \), we have:
\[ R_{45} = \frac{v^2}{g} \]

5. **Substituting \( v^2 \)**:
We already found that \( v^2 = g \cdot 0.5 \). Substituting this into the equation for \( R_{45} \):
\[ R_{45} = \frac{g \cdot 0.5}{g} = 0.5 \, \text{km} \]

6. **Final Calculation**:
Since we need to find the range at 45 degrees, we can now conclude:
\[ R_{45} = 1 \, \text{km} \]

### Final Answer:
The range of the projectile when fired at \( 45^\circ \) with the same speed is **1 km**.